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Solve -2x² - 3x - 5 = 0

-2x² - 3x - 5 = 0 has no real solutions. The two complex solutions are x = (-3 + √31 i)/4 and x = (-3 - √31 i)/4.

x² +x += 0
−2x2−3x−5=0⇒x=−3+31 i4, −3−31 i4 (complex)-2x^2 - 3x - 5 = 0 \Rightarrow x = \frac{-3 + \sqrt{31}\, i}{4},\ \frac{-3 - \sqrt{31}\, i}{4}\ \text{(complex)}
Discriminant D = -31

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

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Graph of y = -2x² - 3x - 5-4-3-2-1123-35-30-25-20-15-10-5vertexy = -2x² - 3x - 5
Graph of y = -2x² - 3x - 5. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.75, -3.875).

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 = -3, c = -5.

    −2x2−3x−5=0-2x^2 - 3x - 5 = 0
  2. 2

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

    D=(−3)2−4⋅(−2)⋅(−5)=9−40=−31D = (-3)^2 - 4\cdot(-2)\cdot(-5) = 9 - 40 = -31
  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±−31−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-31}}{-4}
  5. 5

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

    x=−3+31 i4orx=−3−31 i4x = \frac{-3 + \sqrt{31}\, i}{4}\quad\text{or}\quad x = \frac{-3 - \sqrt{31}\, i}{4}
LaTeX source of the formulas on this page
\[ -2x^2 - 3x - 5 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{31}\, i}{4},\ \frac{-3 - \sqrt{31}\, i}{4} \]
\[ -2x^2 - 3x - 5 = 0 \]
\[ D = (-3)^2 - 4\cdot(-2)\cdot(-5) = 9 - 40 = -31 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-31}}{-4} \]
\[ x = \frac{-3 + \sqrt{31}\, i}{4}\quad\text{or}\quad x = \frac{-3 - \sqrt{31}\, i}{4} \]

The parabola y = -2x² - 3x - 5

xy
-3.75-21.875
-2.75-11.875
-1.75-5.875
-0.75-3.875
0.25-5.875
1.25-11.875
2.25-21.875

Questions

What are the solutions of -2x² - 3x - 5 = 0?

-2x² - 3x - 5 = 0 has no real solutions. The two complex solutions are x = (-3 + √31 i)/4 and x = (-3 - √31 i)/4.

Does -2x² - 3x - 5 = 0 factor?

Not over the integers, because the discriminant -31 is negative. Use the quadratic formula.

What is the vertex of y = -2x² - 3x - 5?

(-3/4, -31/8). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -3/2, product = c/a = 5/2.

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