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Solve for x (complex solution)
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3x^{2}-5x=-10
Subtract 5x from both sides.
3x^{2}-5x+10=0
Add 10 to both sides.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times 10}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times 10}}{2\times 3}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-12\times 10}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-5\right)±\sqrt{25-120}}{2\times 3}
Multiply -12 times 10.
x=\frac{-\left(-5\right)±\sqrt{-95}}{2\times 3}
Add 25 to -120.
x=\frac{-\left(-5\right)±\sqrt{95}i}{2\times 3}
Take the square root of -95.
x=\frac{5±\sqrt{95}i}{2\times 3}
The opposite of -5 is 5.
x=\frac{5±\sqrt{95}i}{6}
Multiply 2 times 3.
x=\frac{5+\sqrt{95}i}{6}
Now solve the equation x=\frac{5±\sqrt{95}i}{6} when ± is plus. Add 5 to i\sqrt{95}.
x=\frac{-\sqrt{95}i+5}{6}
Now solve the equation x=\frac{5±\sqrt{95}i}{6} when ± is minus. Subtract i\sqrt{95} from 5.
x=\frac{5+\sqrt{95}i}{6} x=\frac{-\sqrt{95}i+5}{6}
The equation is now solved.
3x^{2}-5x=-10
Subtract 5x from both sides.
\frac{3x^{2}-5x}{3}=-\frac{10}{3}
Divide both sides by 3.
x^{2}-\frac{5}{3}x=-\frac{10}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{5}{6}\right)^{2}
Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{10}{3}+\frac{25}{36}
Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{95}{36}
Add -\frac{10}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{6}\right)^{2}=-\frac{95}{36}
Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{95}{36}}
Take the square root of both sides of the equation.
x-\frac{5}{6}=\frac{\sqrt{95}i}{6} x-\frac{5}{6}=-\frac{\sqrt{95}i}{6}
Simplify.
x=\frac{5+\sqrt{95}i}{6} x=\frac{-\sqrt{95}i+5}{6}
Add \frac{5}{6} to both sides of the equation.