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Solve -2x² + x - 5 = 0

-2x² + x - 5 = 0 has no real solutions. The two complex solutions are x = (1 + √39 i)/4 and x = (1 - √39 i)/4.

x² +x += 0
−2x2+x−5=0⇒x=1+39 i4, 1−39 i4 (complex)-2x^2 + x - 5 = 0 \Rightarrow x = \frac{1 + \sqrt{39}\, i}{4},\ \frac{1 - \sqrt{39}\, i}{4}\ \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 = -2x² + x - 5-3-2-11234-35-30-25-20-15-10-5vertexy = -2x² + x - 5
Graph of y = -2x² + x - 5. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.25, -4.875).

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 = 1, c = -5.

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

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

    D=12−4⋅(−2)⋅(−5)=1−40=−39D = 1^2 - 4\cdot(-2)\cdot(-5) = 1 - 40 = -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=−1±−39−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{-39}}{-4}
  5. 5

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

    x=1+39 i4orx=1−39 i4x = \frac{1 + \sqrt{39}\, i}{4}\quad\text{or}\quad x = \frac{1 - \sqrt{39}\, i}{4}
LaTeX source of the formulas on this page
\[ -2x^2 + x - 5 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{39}\, i}{4},\ \frac{1 - \sqrt{39}\, i}{4} \]
\[ -2x^2 + x - 5 = 0 \]
\[ D = 1^2 - 4\cdot(-2)\cdot(-5) = 1 - 40 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{-39}}{-4} \]
\[ x = \frac{1 + \sqrt{39}\, i}{4}\quad\text{or}\quad x = \frac{1 - \sqrt{39}\, i}{4} \]

The parabola y = -2x² + x - 5

xy
-2.75-22.875
-1.75-12.875
-0.75-6.875
0.25-4.875
1.25-6.875
2.25-12.875
3.25-22.875

Questions

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

-2x² + x - 5 = 0 has no real solutions. The two complex solutions are x = (1 + √39 i)/4 and x = (1 - √39 i)/4.

Does -2x² + x - 5 = 0 factor?

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

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

(1/4, -39/8). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 1/2, product = c/a = 5/2.

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