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Solve -2x² + 6x - 6 = 0

-2x² + 6x - 6 = 0 has no real solutions. The two complex solutions are x = (3 + √3 i)/2 and x = (3 - √3 i)/2.

x² +x += 0
−2x2+6x−6=0⇒x=3+3 i2, 3−3 i2 (complex)-2x^2 + 6x - 6 = 0 \Rightarrow x = \frac{3 + \sqrt{3}\, i}{2},\ \frac{3 - \sqrt{3}\, i}{2}\ \text{(complex)}
Discriminant D = -12

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

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Graph of y = -2x² + 6x - 6-2-112345-35-30-25-20-15-10-5vertexy = -2x² + 6x - 6
Graph of y = -2x² + 6x - 6. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1.5, -1.5).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -2, b = 6, c = -6.

    −2x2+6x−6=0-2x^2 + 6x - 6 = 0
  2. 2

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

    D=62−4⋅(−2)⋅(−6)=36−48=−12D = 6^2 - 4\cdot(-2)\cdot(-6) = 36 - 48 = -12
  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=−6±−12−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{-12}}{-4}
  5. 5

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

    x=3+3 i2orx=3−3 i2x = \frac{3 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{3}\, i}{2}
LaTeX source of the formulas on this page
\[ -2x^2 + 6x - 6 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{3}\, i}{2},\ \frac{3 - \sqrt{3}\, i}{2} \]
\[ -2x^2 + 6x - 6 = 0 \]
\[ D = 6^2 - 4\cdot(-2)\cdot(-6) = 36 - 48 = -12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{-12}}{-4} \]
\[ x = \frac{3 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{3}\, i}{2} \]

The parabola y = -2x² + 6x - 6

xy
-1.5-19.5
-0.5-9.5
0.5-3.5
1.5-1.5
2.5-3.5
3.5-9.5
4.5-19.5

Questions

What are the solutions of -2x² + 6x - 6 = 0?

-2x² + 6x - 6 = 0 has no real solutions. The two complex solutions are x = (3 + √3 i)/2 and x = (3 - √3 i)/2.

Does -2x² + 6x - 6 = 0 factor?

Not over the integers, because the discriminant -12 is negative. Use the quadratic formula.

What is the vertex of y = -2x² + 6x - 6?

(3/2, -3/2). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 3, product = c/a = 3.

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