Solve 2x² + 11x + 10 = 0
The solutions of 2x² + 11x + 10 = 0 are x = (-11 + √41)/4 or x = (-11 - √41)/4, that is x ≈ -1.1492189 and x ≈ -4.3507811.
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 = 11, 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 ≈ -1.1492189 and x ≈ -4.3507811.
- 7
Check numerically: for x ≈ -1.1492189 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 2x^2 + 11x + 10 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{41}}{4},\ \frac{-11 - \sqrt{41}}{4} \]
\[ 2x^2 + 11x + 10 = 0 \]
\[ D = 11^2 - 4\cdot2\cdot10 = 121 - 80 = 41 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{41}}{4} \]
\[ x = \frac{-11 + \sqrt{41}}{4}\quad\text{or}\quad x = \frac{-11 - \sqrt{41}}{4} \]The parabola y = 2x² + 11x + 10
| x | y |
|---|---|
| -5.75 | 12.875 |
| -4.75 | 2.875 |
| -3.75 | -3.125 |
| -2.75 | -5.125 |
| -1.75 | -3.125 |
| -0.75 | 2.875 |
| 0.25 | 12.875 |
Questions
What are the solutions of 2x² + 11x + 10 = 0?
The solutions of 2x² + 11x + 10 = 0 are x = (-11 + √41)/4 or x = (-11 - √41)/4, that is x ≈ -1.1492189 and x ≈ -4.3507811.
Does 2x² + 11x + 10 = 0 factor?
Not over the integers, because the discriminant 41 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 2x² + 11x + 10?
(-11/4, -41/8). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -11/2, product = c/a = 5.