Solve 3x² + 4x + 12 = 0
3x² + 4x + 12 = 0 has no real solutions. The two complex solutions are x = (-2 + 4√2 i)/3 and x = (-2 - 4√2 i)/3.
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 = 3, b = 4, 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
\[ 3x^2 + 4x + 12 = 0\ \Rightarrow\ x = \frac{-2 + 4\sqrt{2}\, i}{3},\ \frac{-2 - 4\sqrt{2}\, i}{3} \]
\[ 3x^2 + 4x + 12 = 0 \]
\[ D = 4^2 - 4\cdot3\cdot12 = 16 - 144 = -128 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-128}}{6} \]
\[ x = \frac{-2 + 4\sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{-2 - 4\sqrt{2}\, i}{3} \]The parabola y = 3x² + 4x + 12
| x | y |
|---|---|
| -3.6667 | 37.667267 |
| -2.6667 | 22.667067 |
| -1.6667 | 13.666867 |
| -0.6667 | 10.666667 |
| 0.3333 | 13.666467 |
| 1.3333 | 22.666267 |
| 2.3333 | 37.666067 |
Questions
What are the solutions of 3x² + 4x + 12 = 0?
3x² + 4x + 12 = 0 has no real solutions. The two complex solutions are x = (-2 + 4√2 i)/3 and x = (-2 - 4√2 i)/3.
Does 3x² + 4x + 12 = 0 factor?
Not over the integers, because the discriminant -128 is negative. Use the quadratic formula.
What is the vertex of y = 3x² + 4x + 12?
(-2/3, 32/3). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -4/3, product = c/a = 4.