Solve -3x² + 3x - 12 = 0
-3x² + 3x - 12 = 0 has no real solutions. The two complex solutions are x = (1 + √15 i)/2 and x = (1 - √15 i)/2.
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 = -3, 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 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
\[ -3x^2 + 3x - 12 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{15}\, i}{2},\ \frac{1 - \sqrt{15}\, i}{2} \]
\[ -3x^2 + 3x - 12 = 0 \]
\[ D = 3^2 - 4\cdot(-3)\cdot(-12) = 9 - 144 = -135 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-135}}{-6} \]
\[ x = \frac{1 + \sqrt{15}\, i}{2}\quad\text{or}\quad x = \frac{1 - \sqrt{15}\, i}{2} \]The parabola y = -3x² + 3x - 12
| x | y |
|---|---|
| -2.5 | -38.25 |
| -1.5 | -23.25 |
| -0.5 | -14.25 |
| 0.5 | -11.25 |
| 1.5 | -14.25 |
| 2.5 | -23.25 |
| 3.5 | -38.25 |
Questions
What are the solutions of -3x² + 3x - 12 = 0?
-3x² + 3x - 12 = 0 has no real solutions. The two complex solutions are x = (1 + √15 i)/2 and x = (1 - √15 i)/2.
Does -3x² + 3x - 12 = 0 factor?
Not over the integers, because the discriminant -135 is negative. Use the quadratic formula.
What is the vertex of y = -3x² + 3x - 12?
(1/2, -45/4). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 1, product = c/a = 4.