Solve -2x² + 6x + 10 = 0
The solutions of -2x² + 6x + 10 = 0 are x = (3 + √29)/2 or x = (3 - √29)/2, that is x ≈ -1.1925824 and x ≈ 4.1925824.
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 = -2, 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 ≈ -1.1925824 and x ≈ 4.1925824.
- 7
Check numerically: for x ≈ -1.1925824 the left side is 1.7763568e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -2x^2 + 6x + 10 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{29}}{2},\ \frac{3 - \sqrt{29}}{2} \]
\[ -2x^2 + 6x + 10 = 0 \]
\[ D = 6^2 - 4\cdot(-2)\cdot10 = 36 + 80 = 116 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{116}}{-4} \]
\[ x = \frac{3 + \sqrt{29}}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{29}}{2} \]The parabola y = -2x² + 6x + 10
| x | y |
|---|---|
| -1.5 | -3.5 |
| -0.5 | 6.5 |
| 0.5 | 12.5 |
| 1.5 | 14.5 |
| 2.5 | 12.5 |
| 3.5 | 6.5 |
| 4.5 | -3.5 |
Questions
What are the solutions of -2x² + 6x + 10 = 0?
The solutions of -2x² + 6x + 10 = 0 are x = (3 + √29)/2 or x = (3 - √29)/2, that is x ≈ -1.1925824 and x ≈ 4.1925824.
Does -2x² + 6x + 10 = 0 factor?
Not over the integers, because the discriminant 116 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -2x² + 6x + 10?
(3/2, 29/2). 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 = -5.