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Solve -3x² - 11x - 12 = 0

-3x² - 11x - 12 = 0 has no real solutions. The two complex solutions are x = (-11 + √23 i)/6 and x = (-11 - √23 i)/6.

x² +x += 0
−3x2−11x−12=0⇒x=−11+23 i6, −11−23 i6 (complex)-3x^2 - 11x - 12 = 0 \Rightarrow x = \frac{-11 + \sqrt{23}\, i}{6},\ \frac{-11 - \sqrt{23}\, i}{6}\ \text{(complex)}
Discriminant D = -23

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

Graph of y = -3x² - 11x - 12-5-4-3-2-112-50-40-30-20-10vertexy = -3x² - 11x - 12
Graph of y = -3x² - 11x - 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.8333333, -1.9166667).

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 = -11, c = -12.

    −3x2−11x−12=0-3x^2 - 11x - 12 = 0
  2. 2

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

    D=(−11)2−4⋅(−3)⋅(−12)=121−144=−23D = (-11)^2 - 4\cdot(-3)\cdot(-12) = 121 - 144 = -23
  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=11±−23−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{11 \pm \sqrt{-23}}{-6}
  5. 5

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

    x=−11+23 i6orx=−11−23 i6x = \frac{-11 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{-11 - \sqrt{23}\, i}{6}
LaTeX source of the formulas on this page
\[ -3x^2 - 11x - 12 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{23}\, i}{6},\ \frac{-11 - \sqrt{23}\, i}{6} \]
\[ -3x^2 - 11x - 12 = 0 \]
\[ D = (-11)^2 - 4\cdot(-3)\cdot(-12) = 121 - 144 = -23 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{11 \pm \sqrt{-23}}{-6} \]
\[ x = \frac{-11 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{-11 - \sqrt{23}\, i}{6} \]

The parabola y = -3x² - 11x - 12

xy
-4.8333-28.916067
-3.8333-13.916267
-2.8333-4.9164667
-1.8333-1.9166667
-0.8333-4.9168667
0.1667-13.917067
1.1667-28.917267

Questions

What are the solutions of -3x² - 11x - 12 = 0?

-3x² - 11x - 12 = 0 has no real solutions. The two complex solutions are x = (-11 + √23 i)/6 and x = (-11 - √23 i)/6.

Does -3x² - 11x - 12 = 0 factor?

Not over the integers, because the discriminant -23 is negative. Use the quadratic formula.

What is the vertex of y = -3x² - 11x - 12?

(-11/6, -23/12). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -11/3, product = c/a = 4.

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