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Solve -3x² - 6x - 5 = 0

-3x² - 6x - 5 = 0 has no real solutions. The two complex solutions are x = (-3 + √6 i)/3 and x = (-3 - √6 i)/3.

x² +x += 0
−3x2−6x−5=0⇒x=−3+6 i3, −3−6 i3 (complex)-3x^2 - 6x - 5 = 0 \Rightarrow x = \frac{-3 + \sqrt{6}\, i}{3},\ \frac{-3 - \sqrt{6}\, i}{3}\ \text{(complex)}
Discriminant D = -24

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

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

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 = -5.

    −3x2−6x−5=0-3x^2 - 6x - 5 = 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)⋅(−5)=36−60=−24D = (-6)^2 - 4\cdot(-3)\cdot(-5) = 36 - 60 = -24
  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±−24−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-24}}{-6}
  5. 5

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

    x=−3+6 i3orx=−3−6 i3x = \frac{-3 + \sqrt{6}\, i}{3}\quad\text{or}\quad x = \frac{-3 - \sqrt{6}\, i}{3}
LaTeX source of the formulas on this page
\[ -3x^2 - 6x - 5 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{6}\, i}{3},\ \frac{-3 - \sqrt{6}\, i}{3} \]
\[ -3x^2 - 6x - 5 = 0 \]
\[ D = (-6)^2 - 4\cdot(-3)\cdot(-5) = 36 - 60 = -24 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-24}}{-6} \]
\[ x = \frac{-3 + \sqrt{6}\, i}{3}\quad\text{or}\quad x = \frac{-3 - \sqrt{6}\, i}{3} \]

The parabola y = -3x² - 6x - 5

xy
-4-29
-3-14
-2-5
-1-2
0-5
1-14
2-29

Questions

What are the solutions of -3x² - 6x - 5 = 0?

-3x² - 6x - 5 = 0 has no real solutions. The two complex solutions are x = (-3 + √6 i)/3 and x = (-3 - √6 i)/3.

Does -3x² - 6x - 5 = 0 factor?

Not over the integers, because the discriminant -24 is negative. Use the quadratic formula.

What is the vertex of y = -3x² - 6x - 5?

(-1, -2). 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 = 5/3.

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