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Solve -x² + 3x - 10 = 0

-x² + 3x - 10 = 0 has no real solutions. The two complex solutions are x = (3 + √31 i)/2 and x = (3 - √31 i)/2.

x² +x += 0
−x2+3x−10=0⇒x=3+31 i2, 3−31 i2 (complex)-x^2 + 3x - 10 = 0 \Rightarrow x = \frac{3 + \sqrt{31}\, i}{2},\ \frac{3 - \sqrt{31}\, i}{2}\ \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 = -x² + 3x - 10-2-112345-24-22-20-18-16-14-12-10-8-6vertexy = -x² + 3x - 10
Graph of y = -x² + 3x - 10. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1.5, -7.75).

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 = -10.

    −x2+3x−10=0-x^2 + 3x - 10 = 0
  2. 2

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

    D=32−4⋅(−1)⋅(−10)=9−40=−31D = 3^2 - 4\cdot(-1)\cdot(-10) = 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±−31−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-31}}{-2}
  5. 5

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

    x=3+31 i2orx=3−31 i2x = \frac{3 + \sqrt{31}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{31}\, i}{2}
LaTeX source of the formulas on this page
\[ -x^2 + 3x - 10 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{31}\, i}{2},\ \frac{3 - \sqrt{31}\, i}{2} \]
\[ -x^2 + 3x - 10 = 0 \]
\[ D = 3^2 - 4\cdot(-1)\cdot(-10) = 9 - 40 = -31 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-31}}{-2} \]
\[ x = \frac{3 + \sqrt{31}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{31}\, i}{2} \]

The parabola y = -x² + 3x - 10

xy
-1.5-16.75
-0.5-11.75
0.5-8.75
1.5-7.75
2.5-8.75
3.5-11.75
4.5-16.75

Questions

What are the solutions of -x² + 3x - 10 = 0?

-x² + 3x - 10 = 0 has no real solutions. The two complex solutions are x = (3 + √31 i)/2 and x = (3 - √31 i)/2.

Does -x² + 3x - 10 = 0 factor?

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

What is the vertex of y = -x² + 3x - 10?

(3/2, -31/4). 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 = 10.

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