Solve -2x² + 12x + 4 = 0
The solutions of -2x² + 12x + 4 = 0 are x = 3 + √11 or x = 3 - √11, that is x ≈ -0.31662479 and x ≈ 6.3166248.
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 = 4.
- 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 ≈ -0.31662479 and x ≈ 6.3166248.
- 7
Check numerically: for x ≈ -0.31662479 the left side is 4.4408921e-16 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -2x^2 + 12x + 4 = 0\ \Rightarrow\ x = 3 + \sqrt{11},\ 3 - \sqrt{11} \]
\[ -2x^2 + 12x + 4 = 0 \]
\[ D = 12^2 - 4\cdot(-2)\cdot4 = 144 + 32 = 176 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{176}}{-4} \]
\[ x = 3 + \sqrt{11}\quad\text{or}\quad x = 3 - \sqrt{11} \]The parabola y = -2x² + 12x + 4
| x | y |
|---|---|
| 0 | 4 |
| 1 | 14 |
| 2 | 20 |
| 3 | 22 |
| 4 | 20 |
| 5 | 14 |
| 6 | 4 |
Questions
What are the solutions of -2x² + 12x + 4 = 0?
The solutions of -2x² + 12x + 4 = 0 are x = 3 + √11 or x = 3 - √11, that is x ≈ -0.31662479 and x ≈ 6.3166248.
Does -2x² + 12x + 4 = 0 factor?
Not over the integers, because the discriminant 176 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -2x² + 12x + 4?
(3, 22). 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 = -2.