Solve 2x² + 7x + 11 = 0
2x² + 7x + 11 = 0 has no real solutions. The two complex solutions are x = (-7 + √39 i)/4 and x = (-7 - √39 i)/4.
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 = 7, c = 11.
- 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
\[ 2x^2 + 7x + 11 = 0\ \Rightarrow\ x = \frac{-7 + \sqrt{39}\, i}{4},\ \frac{-7 - \sqrt{39}\, i}{4} \]
\[ 2x^2 + 7x + 11 = 0 \]
\[ D = 7^2 - 4\cdot2\cdot11 = 49 - 88 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{-39}}{4} \]
\[ x = \frac{-7 + \sqrt{39}\, i}{4}\quad\text{or}\quad x = \frac{-7 - \sqrt{39}\, i}{4} \]The parabola y = 2x² + 7x + 11
| x | y |
|---|---|
| -4.75 | 22.875 |
| -3.75 | 12.875 |
| -2.75 | 6.875 |
| -1.75 | 4.875 |
| -0.75 | 6.875 |
| 0.25 | 12.875 |
| 1.25 | 22.875 |
Questions
What are the solutions of 2x² + 7x + 11 = 0?
2x² + 7x + 11 = 0 has no real solutions. The two complex solutions are x = (-7 + √39 i)/4 and x = (-7 - √39 i)/4.
Does 2x² + 7x + 11 = 0 factor?
Not over the integers, because the discriminant -39 is negative. Use the quadratic formula.
What is the vertex of y = 2x² + 7x + 11?
(-7/4, 39/8). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -7/2, product = c/a = 11/2.