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Solve for x (complex solution)
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-x^{2}+10x-81=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{-10±\sqrt{10^{2}-4\left(-1\right)\left(-81\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-81\right)}}{2\left(-1\right)}
Square 10.
x=\frac{-10±\sqrt{100+4\left(-81\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-10±\sqrt{100-324}}{2\left(-1\right)}
Multiply 4 times -81.
x=\frac{-10±\sqrt{-224}}{2\left(-1\right)}
Add 100 to -324.
x=\frac{-10±4\sqrt{14}i}{2\left(-1\right)}
Take the square root of -224.
x=\frac{-10±4\sqrt{14}i}{-2}
Multiply 2 times -1.
x=\frac{-10+4\sqrt{14}i}{-2}
Now solve the equation x=\frac{-10±4\sqrt{14}i}{-2} when ± is plus. Add -10 to 4i\sqrt{14}.
x=-2\sqrt{14}i+5
Divide -10+4i\sqrt{14} by -2.
x=\frac{-4\sqrt{14}i-10}{-2}
Now solve the equation x=\frac{-10±4\sqrt{14}i}{-2} when ± is minus. Subtract 4i\sqrt{14} from -10.
x=5+2\sqrt{14}i
Divide -10-4i\sqrt{14} by -2.
x=-2\sqrt{14}i+5 x=5+2\sqrt{14}i
The equation is now solved.
-x^{2}+10x-81=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.
-x^{2}+10x-81-\left(-81\right)=-\left(-81\right)
Add 81 to both sides of the equation.
-x^{2}+10x=-\left(-81\right)
Subtracting -81 from itself leaves 0.
-x^{2}+10x=81
Subtract -81 from 0.
\frac{-x^{2}+10x}{-1}=\frac{81}{-1}
Divide both sides by -1.
x^{2}+\frac{10}{-1}x=\frac{81}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}-10x=\frac{81}{-1}
Divide 10 by -1.
x^{2}-10x=-81
Divide 81 by -1.
x^{2}-10x+\left(-5\right)^{2}=-81+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-10x+25=-81+25
Square -5.
x^{2}-10x+25=-56
Add -81 to 25.
\left(x-5\right)^{2}=-56
Factor x^{2}-10x+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-5\right)^{2}}=\sqrt{-56}
Take the square root of both sides of the equation.
x-5=2\sqrt{14}i x-5=-2\sqrt{14}i
Simplify.
x=5+2\sqrt{14}i x=-2\sqrt{14}i+5
Add 5 to both sides of the equation.