Solve 5x² + 12x - 1 = 0
The solutions of 5x² + 12x - 1 = 0 are x = (-6 + √41)/5 or x = (-6 - √41)/5, that is x ≈ 0.080624847 and x ≈ -2.4806248.
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 = 5, 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 ≈ 0.080624847 and x ≈ -2.4806248.
- 7
Check numerically: for x ≈ 0.080624847 the left side is -4.4408921e-16 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 5x^2 + 12x - 1 = 0\ \Rightarrow\ x = \frac{-6 + \sqrt{41}}{5},\ \frac{-6 - \sqrt{41}}{5} \]
\[ 5x^2 + 12x - 1 = 0 \]
\[ D = 12^2 - 4\cdot5\cdot(-1) = 144 + 20 = 164 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{164}}{10} \]
\[ x = \frac{-6 + \sqrt{41}}{5}\quad\text{or}\quad x = \frac{-6 - \sqrt{41}}{5} \]The parabola y = 5x² + 12x - 1
| x | y |
|---|---|
| -4.2 | 36.8 |
| -3.2 | 11.8 |
| -2.2 | -3.2 |
| -1.2 | -8.2 |
| -0.2 | -3.2 |
| 0.8 | 11.8 |
| 1.8 | 36.8 |
Questions
What are the solutions of 5x² + 12x - 1 = 0?
The solutions of 5x² + 12x - 1 = 0 are x = (-6 + √41)/5 or x = (-6 - √41)/5, that is x ≈ 0.080624847 and x ≈ -2.4806248.
Does 5x² + 12x - 1 = 0 factor?
Not over the integers, because the discriminant 164 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 5x² + 12x - 1?
(-6/5, -41/5). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -12/5, product = c/a = -1/5.