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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.

x² +x += 0
−3x2+4x−12=0⇒x=2+42 i3, 2−42 i3 (complex)-3x^2 + 4x - 12 = 0 \Rightarrow x = \frac{2 + 4\sqrt{2}\, i}{3},\ \frac{2 - 4\sqrt{2}\, i}{3}\ \text{(complex)}
Discriminant D = -128

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

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Graph of y = -3x² + 4x - 12-3-2-11234-60-50-40-30-20-10vertexy = -3x² + 4x - 12
Graph of y = -3x² + 4x - 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.66666667, -10.666667).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -3, b = 4, c = -12.

    −3x2+4x−12=0-3x^2 + 4x - 12 = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=42−4⋅(−3)⋅(−12)=16−144=−128D = 4^2 - 4\cdot(-3)\cdot(-12) = 16 - 144 = -128
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−4±−128−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-128}}{-6}
  5. 5

    Simplify (pull the square out of the root) to get the solutions:

    x=2+42 i3orx=2−42 i3x = \frac{2 + 4\sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{2 - 4\sqrt{2}\, i}{3}
LaTeX source of the formulas on this page
\[ -3x^2 + 4x - 12 = 0\ \Rightarrow\ x = \frac{2 + 4\sqrt{2}\, i}{3},\ \frac{2 - 4\sqrt{2}\, i}{3} \]
\[ -3x^2 + 4x - 12 = 0 \]
\[ D = 4^2 - 4\cdot(-3)\cdot(-12) = 16 - 144 = -128 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-128}}{-6} \]
\[ x = \frac{2 + 4\sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{2 - 4\sqrt{2}\, i}{3} \]

The parabola y = -3x² + 4x - 12

xy
-2.3333-37.666067
-1.3333-22.666267
-0.3333-13.666467
0.6667-10.666667
1.6667-13.666867
2.6667-22.667067
3.6667-37.667267

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 downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 4/3, product = c/a = 4.

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