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Solve for x (complex solution)
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2x^{2}-\sqrt{5}x+2=0
Reorder the terms.
2x^{2}+\left(-\sqrt{5}\right)x+2=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{\left(-\sqrt{5}\right)^{2}-4\times 2\times 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -\sqrt{5} for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{5-4\times 2\times 2}}{2\times 2}
Square -\sqrt{5}.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{5-8\times 2}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{5-16}}{2\times 2}
Multiply -8 times 2.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{-11}}{2\times 2}
Add 5 to -16.
x=\frac{-\left(-\sqrt{5}\right)±\sqrt{11}i}{2\times 2}
Take the square root of -11.
x=\frac{\sqrt{5}±\sqrt{11}i}{2\times 2}
The opposite of -\sqrt{5} is \sqrt{5}.
x=\frac{\sqrt{5}±\sqrt{11}i}{4}
Multiply 2 times 2.
x=\frac{\sqrt{5}+\sqrt{11}i}{4}
Now solve the equation x=\frac{\sqrt{5}±\sqrt{11}i}{4} when ± is plus. Add \sqrt{5} to i\sqrt{11}.
x=\frac{-\sqrt{11}i+\sqrt{5}}{4}
Now solve the equation x=\frac{\sqrt{5}±\sqrt{11}i}{4} when ± is minus. Subtract i\sqrt{11} from \sqrt{5}.
x=\frac{\sqrt{5}+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i+\sqrt{5}}{4}
The equation is now solved.
2x^{2}-x\sqrt{5}=-2
Subtract 2 from both sides. Anything subtracted from zero gives its negation.
2x^{2}-\sqrt{5}x=-2
Reorder the terms.
2x^{2}+\left(-\sqrt{5}\right)x=-2
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\frac{2x^{2}+\left(-\sqrt{5}\right)x}{2}=-\frac{2}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x=-\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x=-1
Divide -2 by 2.
x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x+\left(-\frac{\sqrt{5}}{4}\right)^{2}=-1+\left(-\frac{\sqrt{5}}{4}\right)^{2}
Divide -\frac{\sqrt{5}}{2}, the coefficient of the x term, by 2 to get -\frac{\sqrt{5}}{4}. Then add the square of -\frac{\sqrt{5}}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x+\frac{5}{16}=-1+\frac{5}{16}
Square -\frac{\sqrt{5}}{4}.
x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x+\frac{5}{16}=-\frac{11}{16}
Add -1 to \frac{5}{16}.
\left(x-\frac{\sqrt{5}}{4}\right)^{2}=-\frac{11}{16}
Factor x^{2}+\left(-\frac{\sqrt{5}}{2}\right)x+\frac{5}{16}. 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{\sqrt{5}}{4}\right)^{2}}=\sqrt{-\frac{11}{16}}
Take the square root of both sides of the equation.
x-\frac{\sqrt{5}}{4}=\frac{\sqrt{11}i}{4} x-\frac{\sqrt{5}}{4}=-\frac{\sqrt{11}i}{4}
Simplify.
x=\frac{\sqrt{5}+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i+\sqrt{5}}{4}
Add \frac{\sqrt{5}}{4} to both sides of the equation.