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Solve for x (complex solution)
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x^{2}-5x+6=-6
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^{2}-5x+6-\left(-6\right)=-6-\left(-6\right)
Add 6 to both sides of the equation.
x^{2}-5x+6-\left(-6\right)=0
Subtracting -6 from itself leaves 0.
x^{2}-5x+12=0
Subtract -6 from 6.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 12}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 12}}{2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-48}}{2}
Multiply -4 times 12.
x=\frac{-\left(-5\right)±\sqrt{-23}}{2}
Add 25 to -48.
x=\frac{-\left(-5\right)±\sqrt{23}i}{2}
Take the square root of -23.
x=\frac{5±\sqrt{23}i}{2}
The opposite of -5 is 5.
x=\frac{5+\sqrt{23}i}{2}
Now solve the equation x=\frac{5±\sqrt{23}i}{2} when ± is plus. Add 5 to i\sqrt{23}.
x=\frac{-\sqrt{23}i+5}{2}
Now solve the equation x=\frac{5±\sqrt{23}i}{2} when ± is minus. Subtract i\sqrt{23} from 5.
x=\frac{5+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+5}{2}
The equation is now solved.
x^{2}-5x+6=-6
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.
x^{2}-5x+6-6=-6-6
Subtract 6 from both sides of the equation.
x^{2}-5x=-6-6
Subtracting 6 from itself leaves 0.
x^{2}-5x=-12
Subtract 6 from -6.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-12+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-5x+\frac{25}{4}=-12+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=-\frac{23}{4}
Add -12 to \frac{25}{4}.
\left(x-\frac{5}{2}\right)^{2}=-\frac{23}{4}
Factor x^{2}-5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{\sqrt{23}i}{2} x-\frac{5}{2}=-\frac{\sqrt{23}i}{2}
Simplify.
x=\frac{5+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+5}{2}
Add \frac{5}{2} to both sides of the equation.