Solve -x² + 2x + 12 = 0
The solutions of -x² + 2x + 12 = 0 are x = 1 + √13 or x = 1 - √13, that is x ≈ -2.6055513 and x ≈ 4.6055513.
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 = -1, 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
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ -2.6055513 and x ≈ 4.6055513.
- 7
Check numerically: for x ≈ -2.6055513 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -x^2 + 2x + 12 = 0\ \Rightarrow\ x = 1 + \sqrt{13},\ 1 - \sqrt{13} \]
\[ -x^2 + 2x + 12 = 0 \]
\[ D = 2^2 - 4\cdot(-1)\cdot12 = 4 + 48 = 52 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{52}}{-2} \]
\[ x = 1 + \sqrt{13}\quad\text{or}\quad x = 1 - \sqrt{13} \]The parabola y = -x² + 2x + 12
| x | y |
|---|---|
| -2 | 4 |
| -1 | 9 |
| 0 | 12 |
| 1 | 13 |
| 2 | 12 |
| 3 | 9 |
| 4 | 4 |
Questions
What are the solutions of -x² + 2x + 12 = 0?
The solutions of -x² + 2x + 12 = 0 are x = 1 + √13 or x = 1 - √13, that is x ≈ -2.6055513 and x ≈ 4.6055513.
Does -x² + 2x + 12 = 0 factor?
Not over the integers, because the discriminant 52 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -x² + 2x + 12?
(1, 13). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 2, product = c/a = -12.