Solve -3x² - 6x + 10 = 0
The solutions of -3x² - 6x + 10 = 0 are x = (-3 + √39)/3 or x = (-3 - √39)/3, that is x ≈ -3.081666 and x ≈ 1.081666.
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 = -6, c = 10.
- 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 ≈ -3.081666 and x ≈ 1.081666.
- 7
Check numerically: for x ≈ -3.081666 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -3x^2 - 6x + 10 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{39}}{3},\ \frac{-3 - \sqrt{39}}{3} \]
\[ -3x^2 - 6x + 10 = 0 \]
\[ D = (-6)^2 - 4\cdot(-3)\cdot10 = 36 + 120 = 156 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{156}}{-6} \]
\[ x = \frac{-3 + \sqrt{39}}{3}\quad\text{or}\quad x = \frac{-3 - \sqrt{39}}{3} \]The parabola y = -3x² - 6x + 10
| x | y |
|---|---|
| -4 | -14 |
| -3 | 1 |
| -2 | 10 |
| -1 | 13 |
| 0 | 10 |
| 1 | 1 |
| 2 | -14 |
Questions
What are the solutions of -3x² - 6x + 10 = 0?
The solutions of -3x² - 6x + 10 = 0 are x = (-3 + √39)/3 or x = (-3 - √39)/3, that is x ≈ -3.081666 and x ≈ 1.081666.
Does -3x² - 6x + 10 = 0 factor?
Not over the integers, because the discriminant 156 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -3x² - 6x + 10?
(-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 = -10/3.