Quadratic Formula – Steps to Solve Problems
Use the quadratic formula steps below to solve problems on quadratic equations.
For the free practice problems, please go to the third section of the page.
Using the Quadratic Formula – Steps
Quadratic equations are in this format: ax2 ± bx ± c = 0
Look at the following example of a quadratic equation:
x2 – 4x – 8 = 0
Use the quadratic formula steps below to solve.
Step 1: Coefficients and constants
First of all, identify the coefficients and constants.
“Coefficients” are the a and b variables in the equation. c is a constant.
Our equation is: x2 – 4x – 8 = 0
In our equation a = 1. 1 is the coefficient because: x2 × 1 = x2
The second term of our equation is – 4x, so b = –4.
Finally, the third term of our equation is – 8, so c = –8.
Step 2: Substitute value for b
Substitute the value for b and perform the operations.
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{–4 \pm \sqrt{-4^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{4 \pm \sqrt{16 – 4ac}}{2a}\)
Step 3: Substitute values for a and c
Substitute the values and perform the operations to solve for x.
Remember that a = 1 and c = –8.
\(\text{x} = \frac{4 \pm \sqrt{16 – 4ac}}{2a}\)
\(\text{x} = \frac{4 \pm \sqrt{16 – (4 \times 1 \times -8)}}{2 \times 1}\)
\(\text{x} = \frac{4 \pm \sqrt{16 + 32}}{2}\)
\(\text{x} = \frac{4 \pm \sqrt{48}}{2}\)
Quadratic Formula – Exercises
Instructions: Solve each quadratic equation for x using the quadratic formula. If your answer is not a positive or negative integer, you may leave it as an unsimplified fraction as in the examples above.1) x2 + 13x +36 = 0
2) x2 + 3x – 10 = 0
3) 2x2 – 20x + 32 = 0
4) 3x2 – 6x – 45 = 0
5) 4x2 – 2x – 41 = 0
Quadratic Formula – Answers
1) The answers are: –9 and –4x2 + 13x + 36 = 0
a = 1 b = 13 c = 36
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{-13 \pm \sqrt{13^{2} – (4 \times 1 \times 36)}}{2 \times 1}\)
\(\text{x} = \frac{-13 \pm \sqrt{169 – 144}}{2}\)
\(\text{x} = \frac{-13 \pm \sqrt{25}}{2}\)
\(\text{x} = \frac{-13 \pm 5}{2}\)
\(\text{x} = \frac{-13 – 5}{2} = \frac{-18}{2} = -9 \)
\(\text{x} = \frac{-13 + 5}{2} = \frac{-8}{2} = -4 \)
2) The answers are: –5 and 2
x2 + 3x – 10 = 0
a = 1 b = 3 c = –10
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{-3 \pm \sqrt{3^{2} – (4 \times 1 \times -10)}}{2 \times 1}\)
\(\text{x} = \frac{-3 \pm \sqrt{9 +40}}{2}\)
\(\text{x} = \frac{-3 \pm \sqrt{49}}{2}\)
\(\text{x} = \frac{-3 \pm 7}{2}\)
\(\text{x} = \frac{-3 – 7}{2} = \frac{-10}{2} = -5 \)
\(\text{x} = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \)
3) The answers are: 2 and 8
2x2 – 20x + 32 = 0
a = 2 b = –20 c = 32
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{–20 \pm \sqrt{-20^{2} – (4 \times 2 \times 32)}}{2 \times 2}\)
\(\text{x} = \frac{20 \pm \sqrt{400 – 256}}{4}\)
\(\text{x} = \frac{20 \pm \sqrt{144}}{4}\)
\(\text{x} = \frac{20 \pm 12}{4}\)
\(\text{x} = \frac{20 – 12}{4} = \frac{8}{4} = 2 \)
\(\text{x} = \frac{20 + 12}{4} = \frac{32}{4} = 8 \)
4) The answers are: –3 and 5
3x2 – 6x – 45 = 0
a = 3 b = –6 c = –45
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{–6 \pm \sqrt{-6^{2} – (4 \times 3 \times -45)}}{2 \times 3}\)
\(\text{x} = \frac{6 \pm \sqrt{36 + 540}}{6}\)
\(\text{x} = \frac{6 \pm \sqrt{576}}{6}\)
\(\text{x} = \frac{6 \pm 24}{6}\)
\(\text{x} = \frac{6 – 24}{6} = \frac{-18}{6} = -3 \)
\(\text{x} = \frac{6 + 24}{6} = \frac{30}{6} = 5 \)
5) The answers is: \(\text{x} = \frac{2 \pm 14\sqrt{47}}{8}\)
4x2 – 2x – 41 = 0
a = 4 b = –2 c = –41
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{–2 \pm \sqrt{-2^{2} – (4 \times 4 \times -41)}}{2 \times 4}\)
\(\text{x} = \frac{2 \pm \sqrt{2 + 256}}{8}\)
\(\text{x} = \frac{2 \pm \sqrt{658}}{8}\)
\(\text{x} = \frac{2 \pm 14\sqrt{47}}{8}\)
Format of the Quadratic Formula
You will need to use the quadratic formula on several algebra problems on your math test.
The quadratic formula is in this format:
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
Although the quadratic formula may look confusing and difficult, using it is really quite simple.
All you have to do is identify a, b, and c, and then put in the values from the equation provided.
Then you can solve for x.
Sample Exercise
Use the quadratic formula to solve for x in the following problem:
5x2 + x – 3 = 0
Work out your answer on a piece of paper.
Then check your answer against the solution below.
Answer to the Exercise
Step 1:
Write down the coefficients and constant.
Remember that the basic format of the equation is:
ax2 ± bx ± c = 0
And our equation is: 5x2 + x – 3 = 0
So, in our equation a = 5.
The second term of our equation is x, so b = 1.
Finally, the third term of our equation is – 3, so c = –3.
Step 2:
Substitute the value for b in the quadratic formula and perform the operations.
\(\text{x} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{-1 \pm \sqrt{1^{2} – 4ac}}{2a}\)
\(\text{x} = \frac{-1 \pm \sqrt{1 – 4ac}}{2a}\)
Step 3:
Substitute the values for a and c in the quadratic formula and
perform the operations in order to solve.
Remember that a = 5 and c = –3.
\(\text{x} = \frac{-1 \pm \sqrt{1 – 4ac}}{2a}\)
\(\text{x} = \frac{-1 \pm \sqrt{1 – (4 \times 5 \times -3)}}{2 \times 5}\)
\(\text{x} = \frac{-1 \pm \sqrt{1 + 60}}{10}\)
\(\text{x} = \frac{-1 \pm \sqrt{61}}{10}\)
We look at the quadratic equation in more depth in another post.
You may also want to view the following posts:
Solving Factored Quadratic Equations
