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Solve -3x² - 6x - 12 = 0

-3x² - 6x - 12 = 0 has no real solutions. The two complex solutions are x = -1 + √3 i and x = -1 - √3 i.

x² +x += 0
−3x2−6x−12=0⇒x=−1+3 i, −1−3 i (complex)-3x^2 - 6x - 12 = 0 \Rightarrow x = -1 + \sqrt{3}\, i,\ -1 - \sqrt{3}\, i\ \text{(complex)}
Discriminant D = -108

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

Graph of y = -3x² - 6x - 12-5-4-3-2-1123-60-50-40-30-20-10vertexy = -3x² - 6x - 12
Graph of y = -3x² - 6x - 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1, -9).

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 = -6, c = -12.

    −3x2−6x−12=0-3x^2 - 6x - 12 = 0
  2. 2

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

    D=(−6)2−4⋅(−3)⋅(−12)=36−144=−108D = (-6)^2 - 4\cdot(-3)\cdot(-12) = 36 - 144 = -108
  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±−108−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-108}}{-6}
  5. 5

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

    x=−1+3 iorx=−1−3 ix = -1 + \sqrt{3}\, i\quad\text{or}\quad x = -1 - \sqrt{3}\, i
LaTeX source of the formulas on this page
\[ -3x^2 - 6x - 12 = 0\ \Rightarrow\ x = -1 + \sqrt{3}\, i,\ -1 - \sqrt{3}\, i \]
\[ -3x^2 - 6x - 12 = 0 \]
\[ D = (-6)^2 - 4\cdot(-3)\cdot(-12) = 36 - 144 = -108 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-108}}{-6} \]
\[ x = -1 + \sqrt{3}\, i\quad\text{or}\quad x = -1 - \sqrt{3}\, i \]

The parabola y = -3x² - 6x - 12

xy
-4-36
-3-21
-2-12
-1-9
0-12
1-21
2-36

Questions

What are the solutions of -3x² - 6x - 12 = 0?

-3x² - 6x - 12 = 0 has no real solutions. The two complex solutions are x = -1 + √3 i and x = -1 - √3 i.

Does -3x² - 6x - 12 = 0 factor?

Not over the integers, because the discriminant -108 is negative. Use the quadratic formula.

What is the vertex of y = -3x² - 6x - 12?

(-1, -9). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -2, product = c/a = 4.

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