Solve 3x² + 3x - 10 = 0
The solutions of 3x² + 3x - 10 = 0 are x = (-3 + √129)/6 or x = (-3 - √129)/6, that is x ≈ 1.3929694 and x ≈ -2.3929694.
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 = 3, b = 3, 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 ≈ 1.3929694 and x ≈ -2.3929694.
- 7
Check numerically: for x ≈ 1.3929694 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 3x^2 + 3x - 10 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{129}}{6},\ \frac{-3 - \sqrt{129}}{6} \]
\[ 3x^2 + 3x - 10 = 0 \]
\[ D = 3^2 - 4\cdot3\cdot(-10) = 9 + 120 = 129 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{129}}{6} \]
\[ x = \frac{-3 + \sqrt{129}}{6}\quad\text{or}\quad x = \frac{-3 - \sqrt{129}}{6} \]The parabola y = 3x² + 3x - 10
| x | y |
|---|---|
| -3.5 | 16.25 |
| -2.5 | 1.25 |
| -1.5 | -7.75 |
| -0.5 | -10.75 |
| 0.5 | -7.75 |
| 1.5 | 1.25 |
| 2.5 | 16.25 |
Questions
What are the solutions of 3x² + 3x - 10 = 0?
The solutions of 3x² + 3x - 10 = 0 are x = (-3 + √129)/6 or x = (-3 - √129)/6, that is x ≈ 1.3929694 and x ≈ -2.3929694.
Does 3x² + 3x - 10 = 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 = 3x² + 3x - 10?
(-1/2, -43/4). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -1, product = c/a = -10/3.