Solve 2x² + 4x + 6 = 0
2x² + 4x + 6 = 0 has no real solutions. The two complex solutions are x = -1 + √2 i and x = -1 - √2 i.
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 = 4, c = 6.
- 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
\[ 2x^2 + 4x + 6 = 0\ \Rightarrow\ x = -1 + \sqrt{2}\, i,\ -1 - \sqrt{2}\, i \]
\[ 2x^2 + 4x + 6 = 0 \]
\[ D = 4^2 - 4\cdot2\cdot6 = 16 - 48 = -32 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-32}}{4} \]
\[ x = -1 + \sqrt{2}\, i\quad\text{or}\quad x = -1 - \sqrt{2}\, i \]The parabola y = 2x² + 4x + 6
| x | y |
|---|---|
| -4 | 22 |
| -3 | 12 |
| -2 | 6 |
| -1 | 4 |
| 0 | 6 |
| 1 | 12 |
| 2 | 22 |
Questions
What are the solutions of 2x² + 4x + 6 = 0?
2x² + 4x + 6 = 0 has no real solutions. The two complex solutions are x = -1 + √2 i and x = -1 - √2 i.
Does 2x² + 4x + 6 = 0 factor?
Not over the integers, because the discriminant -32 is negative. Use the quadratic formula.
What is the vertex of y = 2x² + 4x + 6?
(-1, 4). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -2, product = c/a = 3.