Solve -2x² - 12x - 5 = 0
The solutions of -2x² - 12x - 5 = 0 are x = (-6 + √26)/2 or x = (-6 - √26)/2, that is x ≈ -5.5495098 and x ≈ -0.45049024.
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 = -12, c = -5.
- 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 ≈ -5.5495098 and x ≈ -0.45049024.
- 7
Check numerically: for x ≈ -5.5495098 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -2x^2 - 12x - 5 = 0\ \Rightarrow\ x = \frac{-6 + \sqrt{26}}{2},\ \frac{-6 - \sqrt{26}}{2} \]
\[ -2x^2 - 12x - 5 = 0 \]
\[ D = (-12)^2 - 4\cdot(-2)\cdot(-5) = 144 - 40 = 104 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{104}}{-4} \]
\[ x = \frac{-6 + \sqrt{26}}{2}\quad\text{or}\quad x = \frac{-6 - \sqrt{26}}{2} \]The parabola y = -2x² - 12x - 5
| x | y |
|---|---|
| -6 | -5 |
| -5 | 5 |
| -4 | 11 |
| -3 | 13 |
| -2 | 11 |
| -1 | 5 |
| 0 | -5 |
Questions
What are the solutions of -2x² - 12x - 5 = 0?
The solutions of -2x² - 12x - 5 = 0 are x = (-6 + √26)/2 or x = (-6 - √26)/2, that is x ≈ -5.5495098 and x ≈ -0.45049024.
Does -2x² - 12x - 5 = 0 factor?
Not over the integers, because the discriminant 104 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -2x² - 12x - 5?
(-3, 13). 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 = 5/2.