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Solve -x² - 3x - 12 = 0

-x² - 3x - 12 = 0 has no real solutions. The two complex solutions are x = (-3 + √39 i)/2 and x = (-3 - √39 i)/2.

x² +x += 0
−x2−3x−12=0⇒x=−3+39 i2, −3−39 i2 (complex)-x^2 - 3x - 12 = 0 \Rightarrow x = \frac{-3 + \sqrt{39}\, i}{2},\ \frac{-3 - \sqrt{39}\, i}{2}\ \text{(complex)}
Discriminant D = -39

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

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Graph of y = -x² - 3x - 12-5-4-3-2-112-26-24-22-20-18-16-14-12-10-8vertexy = -x² - 3x - 12
Graph of y = -x² - 3x - 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.5, -9.75).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -1, b = -3, c = -12.

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

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

    D=(−3)2−4⋅(−1)⋅(−12)=9−48=−39D = (-3)^2 - 4\cdot(-1)\cdot(-12) = 9 - 48 = -39
  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=3±−39−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-39}}{-2}
  5. 5

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

    x=−3+39 i2orx=−3−39 i2x = \frac{-3 + \sqrt{39}\, i}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{39}\, i}{2}
LaTeX source of the formulas on this page
\[ -x^2 - 3x - 12 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{39}\, i}{2},\ \frac{-3 - \sqrt{39}\, i}{2} \]
\[ -x^2 - 3x - 12 = 0 \]
\[ D = (-3)^2 - 4\cdot(-1)\cdot(-12) = 9 - 48 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-39}}{-2} \]
\[ x = \frac{-3 + \sqrt{39}\, i}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{39}\, i}{2} \]

The parabola y = -x² - 3x - 12

xy
-4.5-18.75
-3.5-13.75
-2.5-10.75
-1.5-9.75
-0.5-10.75
0.5-13.75
1.5-18.75

Questions

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

-x² - 3x - 12 = 0 has no real solutions. The two complex solutions are x = (-3 + √39 i)/2 and x = (-3 - √39 i)/2.

Does -x² - 3x - 12 = 0 factor?

Not over the integers, because the discriminant -39 is negative. Use the quadratic formula.

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

(-3/2, -39/4). 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 = 12.

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