When should I use the quadratic formula instead of factoring?

Try factoring first if the coefficients are small and you spot a factor pair quickly — it is faster. If you have searched for about half a minute without finding factors, or if a is not 1 and the numbers are awkward, switch to the formula. A quick discriminant check settles it: if b^2-4ac is not a perfect square, the equation does not factor over the integers, so factoring will never work.

Do I have to write the equation as equal to zero first?

Yes, always. The formula was derived from ax^2+bx+c=0, so the coefficients only mean what the formula expects when one side is zero. For x^2+3x=10, you must rewrite it as x^2+3x-10=0 and use c=-10. Using c=10 gives a completely different discriminant and wrong solutions.

What does it mean if the discriminant is negative?

It means the equation has no real solutions, and the parabola y=ax^2+bx+c never crosses the x-axis — it stays entirely above or entirely below it. In Algebra 1 you write "no real solutions." In later courses, the square root of a negative number is expressed with imaginary numbers, and the equation turns out to have two complex solutions.

Why do my answers sometimes come out as ugly fractions with radicals?

Because most quadratics simply have irrational roots. An answer like 4(6)5 is exact and complete. Simplify the radical and reduce only by a factor shared by the entire numerator, then stop. Convert to a decimal only when the problem asks for a rounded answer.

ALG1-8.4

The Quadratic Formula & the Discriminant

Master the quadratic formula and the discriminant: solve any quadratic equation step by step and predict 2, 1, or 0 real solutions before you solve.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on The Quadratic Formula & the Discriminant, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start The Quadratic Formula & the Discriminant free

What this lesson covers

Factoring is fast when it works, and square roots handle equations with no middle term — but plenty of quadratics resist both. The quadratic formula never resists: plug in three numbers and it returns every real solution the equation has, rational or irrational. Buried inside it is a smaller expression, the discriminant

b^2-4ac

, that tells you how many real solutions exist before you finish the arithmetic.

In this lesson you will put a quadratic into standard form, identify

a

b

, and

c

with their signs attached, substitute carefully, and simplify the radical. You will also learn to read the discriminant like a preview of the graph: whether the parabola crosses the

x

-axis twice, touches it once, or misses it entirely. These two tools finish the unit's toolkit and get used constantly in later work on quadratic models.

Standard Form and the Formula Itself

The quadratic formula solves any equation written in standard form:

ax^2+bx+c=0,\quad a\neq 0

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The formula is not magic — it is the result of completing the square on the general equation

ax^2+bx+c=0

. Because that work has already been done once, in general, you never have to repeat it for a specific equation.

The single most important habit is this: set the equation equal to zero first. If you are given

2x^2+5x=12

, you cannot use

a=2

b=5

c=12

. Subtract 12 from both sides to get

2x^2+5x-12=0

, so

c=-12

. Students who skip this step get wrong answers even though every later step is done correctly.

Second habit: the coefficients carry their signs. In

3x^2-7x+2=0

b=-7

, not 7. When you substitute,

-b

becomes

-(-7)=7

, and

b^2=(-7)^2=49

, positive. Writing the substitution with parentheses around every coefficient prevents almost all sign errors:

x=\frac{-(-7)\pm\sqrt{(-7)^2-4(3)(2)}}{2(3)}

Third habit: the

\pm

produces two answers, and the entire numerator sits over

2a

. A very common mistake is dividing only the radical part by

2a

, or dropping the

-b

term entirely. Draw the fraction bar long enough to hold the whole numerator, and keep the

\pm

until the final simplification.

Also note the formula works when

b=0

c=0

, though square roots or factoring are usually faster in those cases.

Working Through a Substitution Cleanly

A reliable routine turns the formula into ordinary arithmetic. Solve

2x^2-6x+1=0

.

First, list the coefficients:

a=2

b=-6

c=1

. Second, compute the discriminant separately, off to the side:

b^2-4ac=(-6)^2-4(2)(1)=36-8=28

. Doing this piece on its own is a genuine time-saver, because it is where most arithmetic mistakes happen and because it immediately tells you what kind of answer to expect. Here 28 is positive but not a perfect square, so expect two irrational solutions.

Third, substitute into the formula:

x=\frac{-(-6)\pm\sqrt{28}}{2(2)}=\frac{6\pm\sqrt{28}}{4}

Fourth, simplify the radical:

\sqrt{28}=\sqrt{4\cdot 7}=2\sqrt{7}

, giving

x=\frac{6\pm 2\sqrt{7}}{4}

.

Fifth, reduce. Every term in the numerator and the denominator shares a factor of 2:

x=\frac{2(3\pm\sqrt{7})}{4}=\frac{3\pm\sqrt{7}}{2}

The reduction step is where a specific error shows up repeatedly: you may only cancel a factor common to the entire numerator. Writing

\frac{6\pm 2\sqrt{7}}{4}=\frac{6\pm\sqrt{7}}{2}

is wrong, because only one term was divided. Factoring the numerator first, as shown above, makes the legal cancellation obvious.

If a decimal answer is requested, evaluate both branches separately:

\frac{3+\sqrt7}{2}\approx 2.82

and

\frac{3-\sqrt7}{2}\approx 0.18

. Always check at least one solution by substituting back into the original equation — plugging

x\approx 2.82

into

2x^2-6x+1

gives approximately 0, confirming the work.

The Discriminant: Counting Real Solutions

The discriminant is

D=b^2-4ac

, the expression under the radical. Because the rest of the formula is just

\frac{-b\pm\sqrt{D}}{2a}

, the sign of

D

controls everything about the number of real solutions.

Discriminant	Real solutions	Graph of $y=ax^2+bx+c$	Nature of roots
$D>0$ , perfect square	2	crosses $x$ -axis twice	two rational
$D>0$ , not a perfect square	2	crosses $x$ -axis twice	two irrational
$D=0$	1 (a double root)	vertex sits on the $x$ -axis	one rational
$D<0$	0	never touches the $x$ -axis	no real roots

The reasoning is short. If

D>0

\sqrt{D}

is a positive number, and adding then subtracting it from

-b

gives two different values. If

D=0

\sqrt{0}=0

, so both branches collapse to the single value

x=\frac{-b}{2a}

— which is exactly the

x

-coordinate of the vertex, matching the picture of a parabola tangent to the

x

-axis. If

D<0

, no real number squares to a negative, so the formula produces no real output.

Two cautions. First, the discriminant counts solutions of the equation, which are the

x

-intercepts of the graph, not the number of times the graph appears above the axis. Second, a negative discriminant does not mean "no solution" in some absolute sense — it means no real solution, and later courses extend the number system to handle those cases. In Algebra 1, answer "no real solutions."

A question may ask only for the number of solutions. Then computing

D

is the whole job; do not waste effort finishing the formula.

Choosing a Method and Avoiding the Usual Traps

You now have four ways to solve a quadratic, and part of the skill is picking efficiently.

Situation	Best first choice
No linear term, like $3x^2-27=0$	square roots
Factors quickly over the integers	factoring
$a=1$ and $b$ is even	completing the square
Anything else, or ugly coefficients	quadratic formula

The formula always works, so when you are stuck or short on time, use it. It also gives exact answers, which decimals from a graph cannot.

The traps worth memorizing come from real student work. Forgetting to move every term to one side is the biggest one:

x^2+4x=5

must become

x^2+4x-5=0

before

c=-5

is used. Mis-signing

b

is second: with

b=-3

-b=3

and

b^2=9

. Dividing by

a

instead of

2a

is third — in

5x^2+2x-1=0

the denominator is 10, not 5.

Another frequent slip is treating

-b^2

and

(-b)^2

as the same. In the discriminant you always square the whole coefficient, so

b^2

is never negative. If your discriminant comes out negative, the negativity must come from

-4ac

, which happens when

a

and

c

have the same sign and

|4ac|

exceeds

b^2

.

Finally, present answers correctly. Exact form keeps the radical:

x=\frac{-1\pm\sqrt{6}}{5}

. If the problem says "round to the nearest hundredth," compute both branches separately rather than rounding

\sqrt{6}

too early. And when

D=0

, say there is one real solution (a double root) rather than listing the same number twice as if it were two.

Reading the Discriminant Backward

Some problems reverse the question: instead of giving you all three coefficients, they leave one unknown and tell you how many solutions the equation has. These are solved by translating the sentence into a statement about

D

.

Suppose

x^2+kx+9=0

has exactly one real solution. Exactly one real solution means

D=0

, so

k^2-4(1)(9)=0

, giving

k^2=36

and

k=6

k=-6

. Both values work, and checking confirms it:

x^2+6x+9=(x+3)^2

and

x^2-6x+9=(x-3)^2

, each a perfect square trinomial with the double root

x=-3

x=3

.

Now suppose

2x^2+5x+c=0

has two real solutions. That requires

D>0

, so

25-8c>0

, which gives

c<\frac{25}{8}

. The answer is an inequality, not a single number, because infinitely many values of

c

produce two intercepts.

The same idea connects to graphs. Asking how many times the parabola

y=x^2-4x+m

crosses the

x

-axis is asking for the sign of

16-4m

. It crosses twice when

m<4

, touches once when

m=4

, and misses entirely when

m>4

. Sketching a few of these makes the pattern visible: raising

m

slides the whole parabola upward until it lifts off the axis.

Where students go wrong here is mixing up the inequality direction after dividing by a negative, and assuming "one solution" means

D>0

with a small value. Set up the condition on

D

first, in words, then translate. It also helps to sanity-check a boundary value by substituting it back and computing the discriminant numerically.

Key terms

Standard form of a quadratic.: An equation written as $ax^2+bx+c=0$ with $a\neq 0$ ; the quadratic formula requires this arrangement before coefficients are identified.
Quadratic formula.: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , the solution to every quadratic equation in standard form, derived by completing the square in general.
Discriminant.: The expression $D=b^2-4ac$ under the radical; its sign determines whether there are two, one, or no real solutions.
Double root.: The single solution that occurs when $D=0$ ; the two branches of the formula collapse to $x=\frac{-b}{2a}$ , the vertex's $x$ -coordinate.
Real solution.: A solution that is a real number, corresponding to an $x$ -intercept of the parabola $y=ax^2+bx+c$ .
Exact form.: An answer left with radicals and fractions, such as $\frac{3-\sqrt{7}}{2}$ , rather than rounded to a decimal.
Perfect-square discriminant.: A value of $b^2-4ac$ such as 9, 16, or 49 whose square root is an integer, signaling that the two solutions are rational and the quadratic factors over the integers.

Worked example

Solve

5x^2+2=8x

using the quadratic formula. Give exact solutions in simplest form, then state how many real solutions there are and why.

Step 1 — write standard form. Subtract

8x

from both sides:

5x^2-8x+2=0

. Now

a=5

b=-8

c=2

.

Step 2 — compute the discriminant separately.

D=b^2-4ac=(-8)^2-4(5)(2)=64-40=24

. Since

24>0

and 24 is not a perfect square, expect two irrational real solutions.

Step 3 — substitute into the formula, keeping parentheses around each coefficient.

x=\frac{-(-8)\pm\sqrt{24}}{2(5)}=\frac{8\pm\sqrt{24}}{10}

Step 4 — simplify the radical.

\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}

, so

x=\frac{8\pm 2\sqrt{6}}{10}

.

Step 5 — reduce legally. Factor 2 out of the whole numerator:

x=\frac{2(4\pm\sqrt{6})}{10}=\frac{4\pm\sqrt{6}}{5}

. Note that cancelling from only one term, writing

\frac{8\pm\sqrt6}{5}

, would be wrong.

Step 6 — state and check. The exact solutions are

x=\frac{4+\sqrt{6}}{5}

and

x=\frac{4-\sqrt{6}}{5}

, approximately

1.29

and

0.31

. Checking

x\approx 1.29

5(1.29)^2+2\approx 10.32

and

8(1.29)\approx 10.32

, so the two sides agree.

There are two real solutions because the discriminant 24 is positive, meaning the parabola

y=5x^2-8x+2

crosses the

x

-axis at two distinct points.

Practice questions

How many real solutions does

3x^2-4x+7=0

have?

Two irrational solutions
Two rational solutions
Exactly one real solution
No real solutions

Answer: No real solutions

Identify

a=3

b=-4

c=7

, then compute

D=(-4)^2-4(3)(7)=16-84=-68

. Because the discriminant is negative,

\sqrt{D}

is not a real number, so the formula produces no real output and the parabola

y=3x^2-4x+7

never touches the

x

-axis. A frequent error here is writing

b^2

-16

; squaring

-4

always gives a positive 16.

Solve

x^2+6x-4=0

with the quadratic formula, giving exact simplified answers.

Answer: $x=-3+\sqrt{13}$ and $x=-3-\sqrt{13}$

With

a=1

b=6

c=-4

, the discriminant is

36-4(1)(-4)=36+16=52

. Note the double negative:

-4ac

becomes

+16

because

c

is negative. Then

x=\frac{-6\pm\sqrt{52}}{2}

. Since

\sqrt{52}=2\sqrt{13}

, this is

\frac{-6\pm 2\sqrt{13}}{2}=\frac{2(-3\pm\sqrt{13})}{2}=-3\pm\sqrt{13}

. Approximately, the solutions are

0.61

and

-6.61

For what values of

k

does

x^2-kx+16=0

have exactly one real solution?

Answer: $k=8$ or $k=-8$

Exactly one real solution means the discriminant equals zero. Here

a=1

b=-k

c=16

, so

D=(-k)^2-4(1)(16)=k^2-64

. Setting

k^2-64=0

gives

k^2=64

, so

k=8

k=-8

. Both are valid:

x^2-8x+16=(x-4)^2

has the double root

x=4

, and

x^2+8x+16=(x+4)^2

has the double root

x=-4

. Reporting only

k=8

misses half the answer, because squaring erases the sign.

FAQ

When should I use the quadratic formula instead of factoring?: Try factoring first if the coefficients are small and you spot a factor pair quickly — it is faster. If you have searched for about half a minute without finding factors, or if $a$ is not 1 and the numbers are awkward, switch to the formula. A quick discriminant check settles it: if $b^2-4ac$ is not a perfect square, the equation does not factor over the integers, so factoring will never work.
Do I have to write the equation as equal to zero first?: Yes, always. The formula was derived from $ax^2+bx+c=0$ , so the coefficients only mean what the formula expects when one side is zero. For $x^2+3x=10$ , you must rewrite it as $x^2+3x-10=0$ and use $c=-10$ . Using $c=10$ gives a completely different discriminant and wrong solutions.
What does it mean if the discriminant is negative?: It means the equation has no real solutions, and the parabola $y=ax^2+bx+c$ never crosses the $x$ -axis — it stays entirely above or entirely below it. In Algebra 1 you write "no real solutions." In later courses, the square root of a negative number is expressed with imaginary numbers, and the equation turns out to have two complex solutions.
Why do my answers sometimes come out as ugly fractions with radicals?: Because most quadratics simply have irrational roots. An answer like $\frac{4\pm\sqrt{6}}{5}$ is exact and complete. Simplify the radical and reduce only by a factor shared by the entire numerator, then stop. Convert to a decimal only when the problem asks for a rounded answer.

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