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x^{2}-11x-5=8
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}-11x-5-8=8-8
Subtract 8 from both sides of the equation.
x^{2}-11x-5-8=0
Subtracting 8 from itself leaves 0.
x^{2}-11x-13=0
Subtract 8 from -5.
x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-13\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-11\right)±\sqrt{121-4\left(-13\right)}}{2}
Square -11.
x=\frac{-\left(-11\right)±\sqrt{121+52}}{2}
Multiply -4 times -13.
x=\frac{-\left(-11\right)±\sqrt{173}}{2}
Add 121 to 52.
x=\frac{11±\sqrt{173}}{2}
The opposite of -11 is 11.
x=\frac{\sqrt{173}+11}{2}
Now solve the equation x=\frac{11±\sqrt{173}}{2} when ± is plus. Add 11 to \sqrt{173}.
x=\frac{11-\sqrt{173}}{2}
Now solve the equation x=\frac{11±\sqrt{173}}{2} when ± is minus. Subtract \sqrt{173} from 11.
x=\frac{\sqrt{173}+11}{2} x=\frac{11-\sqrt{173}}{2}
The equation is now solved.
x^{2}-11x-5=8
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}-11x-5-\left(-5\right)=8-\left(-5\right)
Add 5 to both sides of the equation.
x^{2}-11x=8-\left(-5\right)
Subtracting -5 from itself leaves 0.
x^{2}-11x=13
Subtract -5 from 8.
x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=13+\left(-\frac{11}{2}\right)^{2}
Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-11x+\frac{121}{4}=13+\frac{121}{4}
Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-11x+\frac{121}{4}=\frac{173}{4}
Add 13 to \frac{121}{4}.
\left(x-\frac{11}{2}\right)^{2}=\frac{173}{4}
Factor x^{2}-11x+\frac{121}{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{11}{2}\right)^{2}}=\sqrt{\frac{173}{4}}
Take the square root of both sides of the equation.
x-\frac{11}{2}=\frac{\sqrt{173}}{2} x-\frac{11}{2}=-\frac{\sqrt{173}}{2}
Simplify.
x=\frac{\sqrt{173}+11}{2} x=\frac{11-\sqrt{173}}{2}
Add \frac{11}{2} to both sides of the equation.