Solve 5x² + 3x + 10 = 0
5x² + 3x + 10 = 0 has no real solutions. The two complex solutions are x = (-3 + √191 i)/10 and x = (-3 - √191 i)/10.
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 = 10.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is negative, so there are no real solutions; the two solutions are complex numbers.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
LaTeX source of the formulas on this page
\[ 5x^2 + 3x + 10 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{191}\, i}{10},\ \frac{-3 - \sqrt{191}\, i}{10} \]
\[ 5x^2 + 3x + 10 = 0 \]
\[ D = 3^2 - 4\cdot5\cdot10 = 9 - 200 = -191 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-191}}{10} \]
\[ x = \frac{-3 + \sqrt{191}\, i}{10}\quad\text{or}\quad x = \frac{-3 - \sqrt{191}\, i}{10} \]The parabola y = 5x² + 3x + 10
| x | y |
|---|---|
| -3.3 | 54.55 |
| -2.3 | 29.55 |
| -1.3 | 14.55 |
| -0.3 | 9.55 |
| 0.7 | 14.55 |
| 1.7 | 29.55 |
| 2.7 | 54.55 |
Questions
What are the solutions of 5x² + 3x + 10 = 0?
5x² + 3x + 10 = 0 has no real solutions. The two complex solutions are x = (-3 + √191 i)/10 and x = (-3 - √191 i)/10.
Does 5x² + 3x + 10 = 0 factor?
Not over the integers, because the discriminant -191 is negative. Use the quadratic formula.
What is the vertex of y = 5x² + 3x + 10?
(-3/10, 191/20). 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 = 2.