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Solve for x (complex solution)
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5x^{2}-4x=-2
Subtract 4x from both sides.
5x^{2}-4x+2=0
Add 2 to both sides.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\times 2}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\times 2}}{2\times 5}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-20\times 2}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-4\right)±\sqrt{16-40}}{2\times 5}
Multiply -20 times 2.
x=\frac{-\left(-4\right)±\sqrt{-24}}{2\times 5}
Add 16 to -40.
x=\frac{-\left(-4\right)±2\sqrt{6}i}{2\times 5}
Take the square root of -24.
x=\frac{4±2\sqrt{6}i}{2\times 5}
The opposite of -4 is 4.
x=\frac{4±2\sqrt{6}i}{10}
Multiply 2 times 5.
x=\frac{4+2\sqrt{6}i}{10}
Now solve the equation x=\frac{4±2\sqrt{6}i}{10} when ± is plus. Add 4 to 2i\sqrt{6}.
x=\frac{2+\sqrt{6}i}{5}
Divide 4+2i\sqrt{6} by 10.
x=\frac{-2\sqrt{6}i+4}{10}
Now solve the equation x=\frac{4±2\sqrt{6}i}{10} when ± is minus. Subtract 2i\sqrt{6} from 4.
x=\frac{-\sqrt{6}i+2}{5}
Divide 4-2i\sqrt{6} by 10.
x=\frac{2+\sqrt{6}i}{5} x=\frac{-\sqrt{6}i+2}{5}
The equation is now solved.
5x^{2}-4x=-2
Subtract 4x from both sides.
\frac{5x^{2}-4x}{5}=-\frac{2}{5}
Divide both sides by 5.
x^{2}-\frac{4}{5}x=-\frac{2}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-\frac{2}{5}+\left(-\frac{2}{5}\right)^{2}
Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{2}{5}+\frac{4}{25}
Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{6}{25}
Add -\frac{2}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{2}{5}\right)^{2}=-\frac{6}{25}
Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. 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{2}{5}\right)^{2}}=\sqrt{-\frac{6}{25}}
Take the square root of both sides of the equation.
x-\frac{2}{5}=\frac{\sqrt{6}i}{5} x-\frac{2}{5}=-\frac{\sqrt{6}i}{5}
Simplify.
x=\frac{2+\sqrt{6}i}{5} x=\frac{-\sqrt{6}i+2}{5}
Add \frac{2}{5} to both sides of the equation.