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Solve for x (complex solution)
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5x^{2}-6x=-7
Subtract 6x from both sides.
5x^{2}-6x+7=0
Add 7 to both sides.
x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\times 7}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -6 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\times 5\times 7}}{2\times 5}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36-20\times 7}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-6\right)±\sqrt{36-140}}{2\times 5}
Multiply -20 times 7.
x=\frac{-\left(-6\right)±\sqrt{-104}}{2\times 5}
Add 36 to -140.
x=\frac{-\left(-6\right)±2\sqrt{26}i}{2\times 5}
Take the square root of -104.
x=\frac{6±2\sqrt{26}i}{2\times 5}
The opposite of -6 is 6.
x=\frac{6±2\sqrt{26}i}{10}
Multiply 2 times 5.
x=\frac{6+2\sqrt{26}i}{10}
Now solve the equation x=\frac{6±2\sqrt{26}i}{10} when ± is plus. Add 6 to 2i\sqrt{26}.
x=\frac{3+\sqrt{26}i}{5}
Divide 6+2i\sqrt{26} by 10.
x=\frac{-2\sqrt{26}i+6}{10}
Now solve the equation x=\frac{6±2\sqrt{26}i}{10} when ± is minus. Subtract 2i\sqrt{26} from 6.
x=\frac{-\sqrt{26}i+3}{5}
Divide 6-2i\sqrt{26} by 10.
x=\frac{3+\sqrt{26}i}{5} x=\frac{-\sqrt{26}i+3}{5}
The equation is now solved.
5x^{2}-6x=-7
Subtract 6x from both sides.
\frac{5x^{2}-6x}{5}=-\frac{7}{5}
Divide both sides by 5.
x^{2}-\frac{6}{5}x=-\frac{7}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=-\frac{7}{5}+\left(-\frac{3}{5}\right)^{2}
Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{7}{5}+\frac{9}{25}
Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{26}{25}
Add -\frac{7}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{5}\right)^{2}=-\frac{26}{25}
Factor x^{2}-\frac{6}{5}x+\frac{9}{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{3}{5}\right)^{2}}=\sqrt{-\frac{26}{25}}
Take the square root of both sides of the equation.
x-\frac{3}{5}=\frac{\sqrt{26}i}{5} x-\frac{3}{5}=-\frac{\sqrt{26}i}{5}
Simplify.
x=\frac{3+\sqrt{26}i}{5} x=\frac{-\sqrt{26}i+3}{5}
Add \frac{3}{5} to both sides of the equation.