Solve 5x² + 3x - 6 = 0
The solutions of 5x² + 3x - 6 = 0 are x = (-3 + √129)/10 or x = (-3 - √129)/10, that is x ≈ 0.83578167 and x ≈ -1.4357817.
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 = 3, c = -6.
- 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 ≈ 0.83578167 and x ≈ -1.4357817.
- 7
Check numerically: for x ≈ 0.83578167 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 5x^2 + 3x - 6 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{129}}{10},\ \frac{-3 - \sqrt{129}}{10} \]
\[ 5x^2 + 3x - 6 = 0 \]
\[ D = 3^2 - 4\cdot5\cdot(-6) = 9 + 120 = 129 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{129}}{10} \]
\[ x = \frac{-3 + \sqrt{129}}{10}\quad\text{or}\quad x = \frac{-3 - \sqrt{129}}{10} \]The parabola y = 5x² + 3x - 6
| x | y |
|---|---|
| -3.3 | 38.55 |
| -2.3 | 13.55 |
| -1.3 | -1.45 |
| -0.3 | -6.45 |
| 0.7 | -1.45 |
| 1.7 | 13.55 |
| 2.7 | 38.55 |
Questions
What are the solutions of 5x² + 3x - 6 = 0?
The solutions of 5x² + 3x - 6 = 0 are x = (-3 + √129)/10 or x = (-3 - √129)/10, that is x ≈ 0.83578167 and x ≈ -1.4357817.
Does 5x² + 3x - 6 = 0 factor?
Not over the integers, because the discriminant 129 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 5x² + 3x - 6?
(-3/10, -129/20). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -3/5, product = c/a = -6/5.