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Solve for x (complex solution)
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x^{2}-3x+5=0
Use the distributive property to multiply x by x-3.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 5}}{2}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-20}}{2}
Multiply -4 times 5.
x=\frac{-\left(-3\right)±\sqrt{-11}}{2}
Add 9 to -20.
x=\frac{-\left(-3\right)±\sqrt{11}i}{2}
Take the square root of -11.
x=\frac{3±\sqrt{11}i}{2}
The opposite of -3 is 3.
x=\frac{3+\sqrt{11}i}{2}
Now solve the equation x=\frac{3±\sqrt{11}i}{2} when ± is plus. Add 3 to i\sqrt{11}.
x=\frac{-\sqrt{11}i+3}{2}
Now solve the equation x=\frac{3±\sqrt{11}i}{2} when ± is minus. Subtract i\sqrt{11} from 3.
x=\frac{3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+3}{2}
The equation is now solved.
x^{2}-3x+5=0
Use the distributive property to multiply x by x-3.
x^{2}-3x=-5
Subtract 5 from both sides. Anything subtracted from zero gives its negation.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-5+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-3x+\frac{9}{4}=-5+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-3x+\frac{9}{4}=-\frac{11}{4}
Add -5 to \frac{9}{4}.
\left(x-\frac{3}{2}\right)^{2}=-\frac{11}{4}
Factor x^{2}-3x+\frac{9}{4}. 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}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{\sqrt{11}i}{2} x-\frac{3}{2}=-\frac{\sqrt{11}i}{2}
Simplify.
x=\frac{3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+3}{2}
Add \frac{3}{2} to both sides of the equation.