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x^{2}+6x=11
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}+6x-11=11-11
Subtract 11 from both sides of the equation.
x^{2}+6x-11=0
Subtracting 11 from itself leaves 0.
x=\frac{-6±\sqrt{6^{2}-4\left(-11\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\left(-11\right)}}{2}
Square 6.
x=\frac{-6±\sqrt{36+44}}{2}
Multiply -4 times -11.
x=\frac{-6±\sqrt{80}}{2}
Add 36 to 44.
x=\frac{-6±4\sqrt{5}}{2}
Take the square root of 80.
x=\frac{4\sqrt{5}-6}{2}
Now solve the equation x=\frac{-6±4\sqrt{5}}{2} when ± is plus. Add -6 to 4\sqrt{5}.
x=2\sqrt{5}-3
Divide -6+4\sqrt{5} by 2.
x=\frac{-4\sqrt{5}-6}{2}
Now solve the equation x=\frac{-6±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from -6.
x=-2\sqrt{5}-3
Divide -6-4\sqrt{5} by 2.
x=2\sqrt{5}-3 x=-2\sqrt{5}-3
The equation is now solved.
x^{2}+6x=11
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}+6x+3^{2}=11+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+6x+9=11+9
Square 3.
x^{2}+6x+9=20
Add 11 to 9.
\left(x+3\right)^{2}=20
Factor x^{2}+6x+9. 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+3\right)^{2}}=\sqrt{20}
Take the square root of both sides of the equation.
x+3=2\sqrt{5} x+3=-2\sqrt{5}
Simplify.
x=2\sqrt{5}-3 x=-2\sqrt{5}-3
Subtract 3 from both sides of the equation.