Solve 2x² + 12x - 1 = 0
The solutions of 2x² + 12x - 1 = 0 are x = (-6 + √38)/2 or x = (-6 - √38)/2, that is x ≈ 0.082207001 and x ≈ -6.082207.
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 = -1.
- 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 ≈ 0.082207001 and x ≈ -6.082207.
- 7
Check numerically: for x ≈ 0.082207001 the left side is -2.6645353e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 2x^2 + 12x - 1 = 0\ \Rightarrow\ x = \frac{-6 + \sqrt{38}}{2},\ \frac{-6 - \sqrt{38}}{2} \]
\[ 2x^2 + 12x - 1 = 0 \]
\[ D = 12^2 - 4\cdot2\cdot(-1) = 144 + 8 = 152 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{152}}{4} \]
\[ x = \frac{-6 + \sqrt{38}}{2}\quad\text{or}\quad x = \frac{-6 - \sqrt{38}}{2} \]The parabola y = 2x² + 12x - 1
| x | y |
|---|---|
| -6 | -1 |
| -5 | -11 |
| -4 | -17 |
| -3 | -19 |
| -2 | -17 |
| -1 | -11 |
| 0 | -1 |
Questions
What are the solutions of 2x² + 12x - 1 = 0?
The solutions of 2x² + 12x - 1 = 0 are x = (-6 + √38)/2 or x = (-6 - √38)/2, that is x ≈ 0.082207001 and x ≈ -6.082207.
Does 2x² + 12x - 1 = 0 factor?
Not over the integers, because the discriminant 152 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 2x² + 12x - 1?
(-3, -19). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -6, product = c/a = -1/2.