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Solve -3x² + 3x - 10 = 0

-3x² + 3x - 10 = 0 has no real solutions. The two complex solutions are x = (3 + √111 i)/6 and x = (3 - √111 i)/6.

x² +x += 0
−3x2+3x−10=0⇒x=3+111 i6, 3−111 i6 (complex)-3x^2 + 3x - 10 = 0 \Rightarrow x = \frac{3 + \sqrt{111}\, i}{6},\ \frac{3 - \sqrt{111}\, i}{6}\ \text{(complex)}
Discriminant D = -111

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

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Graph of y = -3x² + 3x - 10-3-2-11234-60-50-40-30-20-10vertexy = -3x² + 3x - 10
Graph of y = -3x² + 3x - 10. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.5, -9.25).

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 = 3, c = -10.

    −3x2+3x−10=0-3x^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⋅(−3)⋅(−10)=9−120=−111D = 3^2 - 4\cdot(-3)\cdot(-10) = 9 - 120 = -111
  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±−111−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-111}}{-6}
  5. 5

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

    x=3+111 i6orx=3−111 i6x = \frac{3 + \sqrt{111}\, i}{6}\quad\text{or}\quad x = \frac{3 - \sqrt{111}\, i}{6}
LaTeX source of the formulas on this page
\[ -3x^2 + 3x - 10 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{111}\, i}{6},\ \frac{3 - \sqrt{111}\, i}{6} \]
\[ -3x^2 + 3x - 10 = 0 \]
\[ D = 3^2 - 4\cdot(-3)\cdot(-10) = 9 - 120 = -111 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-111}}{-6} \]
\[ x = \frac{3 + \sqrt{111}\, i}{6}\quad\text{or}\quad x = \frac{3 - \sqrt{111}\, i}{6} \]

The parabola y = -3x² + 3x - 10

xy
-2.5-36.25
-1.5-21.25
-0.5-12.25
0.5-9.25
1.5-12.25
2.5-21.25
3.5-36.25

Questions

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

-3x² + 3x - 10 = 0 has no real solutions. The two complex solutions are x = (3 + √111 i)/6 and x = (3 - √111 i)/6.

Does -3x² + 3x - 10 = 0 factor?

Not over the integers, because the discriminant -111 is negative. Use the quadratic formula.

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

(1/2, -37/4). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 1, product = c/a = 10/3.

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