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Solve for x (complex solution)
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2x^{2}-6x+5=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(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 5}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 5}}{2\times 2}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36-8\times 5}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-6\right)±\sqrt{36-40}}{2\times 2}
Multiply -8 times 5.
x=\frac{-\left(-6\right)±\sqrt{-4}}{2\times 2}
Add 36 to -40.
x=\frac{-\left(-6\right)±2i}{2\times 2}
Take the square root of -4.
x=\frac{6±2i}{2\times 2}
The opposite of -6 is 6.
x=\frac{6±2i}{4}
Multiply 2 times 2.
x=\frac{6+2i}{4}
Now solve the equation x=\frac{6±2i}{4} when ± is plus. Add 6 to 2i.
x=\frac{3}{2}+\frac{1}{2}i
Divide 6+2i by 4.
x=\frac{6-2i}{4}
Now solve the equation x=\frac{6±2i}{4} when ± is minus. Subtract 2i from 6.
x=\frac{3}{2}-\frac{1}{2}i
Divide 6-2i by 4.
x=\frac{3}{2}+\frac{1}{2}i x=\frac{3}{2}-\frac{1}{2}i
The equation is now solved.
2x^{2}-6x+5=0
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.
2x^{2}-6x+5-5=-5
Subtract 5 from both sides of the equation.
2x^{2}-6x=-5
Subtracting 5 from itself leaves 0.
\frac{2x^{2}-6x}{2}=-\frac{5}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{6}{2}\right)x=-\frac{5}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-3x=-\frac{5}{2}
Divide -6 by 2.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-\frac{5}{2}+\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}=-\frac{5}{2}+\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{1}{4}
Add -\frac{5}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{2}\right)^{2}=-\frac{1}{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{1}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{1}{2}i x-\frac{3}{2}=-\frac{1}{2}i
Simplify.
x=\frac{3}{2}+\frac{1}{2}i x=\frac{3}{2}-\frac{1}{2}i
Add \frac{3}{2} to both sides of the equation.