Solve -3x² + 4x + 12 = 0
The solutions of -3x² + 4x + 12 = 0 are x = (2 + 2√10)/3 or x = (2 - 2√10)/3, that is x ≈ -1.4415184 and x ≈ 2.7748518.
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 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 ≈ -1.4415184 and x ≈ 2.7748518.
- 7
Check numerically: for x ≈ -1.4415184 the left side is 0 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -3x^2 + 4x + 12 = 0\ \Rightarrow\ x = \frac{2 + 2\sqrt{10}}{3},\ \frac{2 - 2\sqrt{10}}{3} \]
\[ -3x^2 + 4x + 12 = 0 \]
\[ D = 4^2 - 4\cdot(-3)\cdot12 = 16 + 144 = 160 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{160}}{-6} \]
\[ x = \frac{2 + 2\sqrt{10}}{3}\quad\text{or}\quad x = \frac{2 - 2\sqrt{10}}{3} \]The parabola y = -3x² + 4x + 12
| x | y |
|---|---|
| -2.3333 | -13.666067 |
| -1.3333 | 1.3337333 |
| -0.3333 | 10.333533 |
| 0.6667 | 13.333333 |
| 1.6667 | 10.333133 |
| 2.6667 | 1.3329333 |
| 3.6667 | -13.667267 |
Questions
What are the solutions of -3x² + 4x + 12 = 0?
The solutions of -3x² + 4x + 12 = 0 are x = (2 + 2√10)/3 or x = (2 - 2√10)/3, that is x ≈ -1.4415184 and x ≈ 2.7748518.
Does -3x² + 4x + 12 = 0 factor?
Not over the integers, because the discriminant 160 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -3x² + 4x + 12?
(2/3, 40/3). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 4/3, product = c/a = -4.