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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.

Graph of y = 2x² + 6x + 6-5-4-3-2-1125101520253035vertexy = 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=02x^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\cdot2\cdot6 = 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±−124x = \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
\[ 2x^2 + 6x + 6 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{3}\, i}{2},\ \frac{-3 - \sqrt{3}\, i}{2} \]
\[ 2x^2 + 6x + 6 = 0 \]
\[ D = 6^2 - 4\cdot2\cdot6 = 36 - 48 = -12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{-12}}{4} \]
\[ x = \frac{-3 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{3}\, i}{2} \]

The parabola y = 2x² + 6x + 6

xy
-4.519.5
-3.59.5
-2.53.5
-1.51.5
-0.53.5
0.59.5
1.519.5

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 upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -3, product = c/a = 3.

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