Solve x² + 6x + 12 = 0
x² + 6x + 12 = 0 has no real solutions. The two complex solutions are x = -3 + √3 i and x = -3 - √3 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 = 1, b = 6, c = 12.
- 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
\[ x^2 + 6x + 12 = 0\ \Rightarrow\ x = -3 + \sqrt{3}\, i,\ -3 - \sqrt{3}\, i \]
\[ x^2 + 6x + 12 = 0 \]
\[ D = 6^2 - 4\cdot1\cdot12 = 36 - 48 = -12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{-12}}{2} \]
\[ x = -3 + \sqrt{3}\, i\quad\text{or}\quad x = -3 - \sqrt{3}\, i \]The parabola y = x² + 6x + 12
| x | y |
|---|---|
| -6 | 12 |
| -5 | 7 |
| -4 | 4 |
| -3 | 3 |
| -2 | 4 |
| -1 | 7 |
| 0 | 12 |
Questions
What are the solutions of x² + 6x + 12 = 0?
x² + 6x + 12 = 0 has no real solutions. The two complex solutions are x = -3 + √3 i and x = -3 - √3 i.
Does x² + 6x + 12 = 0 factor?
Not over the integers, because the discriminant -12 is negative. Use the quadratic formula.
What is the vertex of y = x² + 6x + 12?
(-3, 3). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -6, product = c/a = 12.