Solve -3x² + 2x - 6 = 0
-3x² + 2x - 6 = 0 has no real solutions. The two complex solutions are x = (1 + √17 i)/3 and x = (1 - √17 i)/3.
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 = 2, c = -6.
- 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
\[ -3x^2 + 2x - 6 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{17}\, i}{3},\ \frac{1 - \sqrt{17}\, i}{3} \]
\[ -3x^2 + 2x - 6 = 0 \]
\[ D = 2^2 - 4\cdot(-3)\cdot(-6) = 4 - 72 = -68 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-68}}{-6} \]
\[ x = \frac{1 + \sqrt{17}\, i}{3}\quad\text{or}\quad x = \frac{1 - \sqrt{17}\, i}{3} \]The parabola y = -3x² + 2x - 6
| x | y |
|---|---|
| -2.6667 | -32.667267 |
| -1.6667 | -17.667067 |
| -0.6667 | -8.6668667 |
| 0.3333 | -5.6666667 |
| 1.3333 | -8.6664667 |
| 2.3333 | -17.666267 |
| 3.3333 | -32.666067 |
Questions
What are the solutions of -3x² + 2x - 6 = 0?
-3x² + 2x - 6 = 0 has no real solutions. The two complex solutions are x = (1 + √17 i)/3 and x = (1 - √17 i)/3.
Does -3x² + 2x - 6 = 0 factor?
Not over the integers, because the discriminant -68 is negative. Use the quadratic formula.
What is the vertex of y = -3x² + 2x - 6?
(1/3, -17/3). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 2/3, product = c/a = 2.