Solve -2x² + 12x + 7 = 0
The solutions of -2x² + 12x + 7 = 0 are x = (6 + 5√2)/2 or x = (6 - 5√2)/2, that is x ≈ -0.53553391 and x ≈ 6.5355339.
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 = 12, c = 7.
- 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 ≈ -0.53553391 and x ≈ 6.5355339.
- 7
Check numerically: for x ≈ -0.53553391 the left side is -1.7763568e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -2x^2 + 12x + 7 = 0\ \Rightarrow\ x = \frac{6 + 5\sqrt{2}}{2},\ \frac{6 - 5\sqrt{2}}{2} \]
\[ -2x^2 + 12x + 7 = 0 \]
\[ D = 12^2 - 4\cdot(-2)\cdot7 = 144 + 56 = 200 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{200}}{-4} \]
\[ x = \frac{6 + 5\sqrt{2}}{2}\quad\text{or}\quad x = \frac{6 - 5\sqrt{2}}{2} \]The parabola y = -2x² + 12x + 7
| x | y |
|---|---|
| 0 | 7 |
| 1 | 17 |
| 2 | 23 |
| 3 | 25 |
| 4 | 23 |
| 5 | 17 |
| 6 | 7 |
Questions
What are the solutions of -2x² + 12x + 7 = 0?
The solutions of -2x² + 12x + 7 = 0 are x = (6 + 5√2)/2 or x = (6 - 5√2)/2, that is x ≈ -0.53553391 and x ≈ 6.5355339.
Does -2x² + 12x + 7 = 0 factor?
Not over the integers, because the discriminant 200 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -2x² + 12x + 7?
(3, 25). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 6, product = c/a = -7/2.