Solve 2x² + 2x - 12 = 0
The solutions of 2x² + 2x - 12 = 0 are x = 2 or x = -3.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 2, b = 2, c = -12.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
Method 1, factoring. The roots are rational, so the left side factors:
- 5
A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:
- 6
Method 2, The quadratic formula works for every quadratic:
- 7
Simplify to get the solutions:
- 8
Check by putting x = 2 back in: a·x² + b·x + c = 2·(4) + 2·(2) + (-12) = 0 ✓
LaTeX source of the formulas on this page
\[ 2x^2 + 2x - 12 = 0\ \Rightarrow\ x = 2,\ -3 \]
\[ 2x^2 + 2x - 12 = 0 \]
\[ D = 2^2 - 4\cdot2\cdot(-12) = 4 + 96 = 100 \]
\[ 2\left(x - 2\right)\left(x + 3\right) = 0 \]
\[ x - 2 = 0 \Rightarrow x = 2\qquad x + 3 = 0 \Rightarrow x = -3 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{100}}{4} \]
\[ x = 2\quad\text{or}\quad x = -3 \]The parabola y = 2x² + 2x - 12
| x | y |
|---|---|
| -3.5 | 5.5 |
| -2.5 | -4.5 |
| -1.5 | -10.5 |
| -0.5 | -12.5 |
| 0.5 | -10.5 |
| 1.5 | -4.5 |
| 2.5 | 5.5 |
Questions
What are the solutions of 2x² + 2x - 12 = 0?
The solutions of 2x² + 2x - 12 = 0 are x = 2 or x = -3.
Does 2x² + 2x - 12 = 0 factor?
Yes: 2(x-2)(x+3) = 0.
What is the vertex of y = 2x² + 2x - 12?
(-1/2, -25/2). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -1, product = c/a = -6.