Solve 2x² + 6x - 12 = 0
The solutions of 2x² + 6x - 12 = 0 are x = (-3 + √33)/2 or x = (-3 - √33)/2, that is x ≈ 1.3722813 and x ≈ -4.3722813.
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 = 6, 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
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ 1.3722813 and x ≈ -4.3722813.
- 7
Check numerically: for x ≈ 1.3722813 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 2x^2 + 6x - 12 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{33}}{2},\ \frac{-3 - \sqrt{33}}{2} \]
\[ 2x^2 + 6x - 12 = 0 \]
\[ D = 6^2 - 4\cdot2\cdot(-12) = 36 + 96 = 132 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{132}}{4} \]
\[ x = \frac{-3 + \sqrt{33}}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{33}}{2} \]The parabola y = 2x² + 6x - 12
| x | y |
|---|---|
| -4.5 | 1.5 |
| -3.5 | -8.5 |
| -2.5 | -14.5 |
| -1.5 | -16.5 |
| -0.5 | -14.5 |
| 0.5 | -8.5 |
| 1.5 | 1.5 |
Questions
What are the solutions of 2x² + 6x - 12 = 0?
The solutions of 2x² + 6x - 12 = 0 are x = (-3 + √33)/2 or x = (-3 - √33)/2, that is x ≈ 1.3722813 and x ≈ -4.3722813.
Does 2x² + 6x - 12 = 0 factor?
Not over the integers, because the discriminant 132 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 2x² + 6x - 12?
(-3/2, -33/2). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -3, product = c/a = -6.