Solve -x² + 6x - 12 = 0
-x² + 6x - 12 = 0 has no real solutions. The two complex solutions are x = 3 + √3 i and x = 3 - √3 i.
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 = 6, c = -12.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is negative, so there are no real solutions; the two solutions are complex numbers.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
LaTeX source of the formulas on this page
\[ -x^2 + 6x - 12 = 0\ \Rightarrow\ x = 3 + \sqrt{3}\, i,\ 3 - \sqrt{3}\, i \]
\[ -x^2 + 6x - 12 = 0 \]
\[ D = 6^2 - 4\cdot(-1)\cdot(-12) = 36 - 48 = -12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{-12}}{-2} \]
\[ x = 3 + \sqrt{3}\, i\quad\text{or}\quad x = 3 - \sqrt{3}\, i \]The parabola y = -x² + 6x - 12
| x | y |
|---|---|
| 0 | -12 |
| 1 | -7 |
| 2 | -4 |
| 3 | -3 |
| 4 | -4 |
| 5 | -7 |
| 6 | -12 |
Questions
What are the solutions of -x² + 6x - 12 = 0?
-x² + 6x - 12 = 0 has no real solutions. The two complex solutions are x = 3 + √3 i and x = 3 - √3 i.
Does -x² + 6x - 12 = 0 factor?
Not over the integers, because the discriminant -12 is negative. Use the quadratic formula.
What is the vertex of y = -x² + 6x - 12?
(3, -3). 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 = 12.