Solve -x² + 3x + 12 = 0
The solutions of -x² + 3x + 12 = 0 are x = (3 + √57)/2 or x = (3 - √57)/2, that is x ≈ -2.2749172 and x ≈ 5.2749172.
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 = 3, 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 ≈ -2.2749172 and x ≈ 5.2749172.
- 7
Check numerically: for x ≈ -2.2749172 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -x^2 + 3x + 12 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{57}}{2},\ \frac{3 - \sqrt{57}}{2} \]
\[ -x^2 + 3x + 12 = 0 \]
\[ D = 3^2 - 4\cdot(-1)\cdot12 = 9 + 48 = 57 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{57}}{-2} \]
\[ x = \frac{3 + \sqrt{57}}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{57}}{2} \]The parabola y = -x² + 3x + 12
| x | y |
|---|---|
| -1.5 | 5.25 |
| -0.5 | 10.25 |
| 0.5 | 13.25 |
| 1.5 | 14.25 |
| 2.5 | 13.25 |
| 3.5 | 10.25 |
| 4.5 | 5.25 |
Questions
What are the solutions of -x² + 3x + 12 = 0?
The solutions of -x² + 3x + 12 = 0 are x = (3 + √57)/2 or x = (3 - √57)/2, that is x ≈ -2.2749172 and x ≈ 5.2749172.
Does -x² + 3x + 12 = 0 factor?
Not over the integers, because the discriminant 57 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -x² + 3x + 12?
(3/2, 57/4). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 3, product = c/a = -12.