Solve for x
x=12
x=0
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x^{2}-12x+36=36
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.
x^{2}-12x+36-36=0
Subtract 36 from both sides.
x^{2}-12x=0
Subtract 36 from 36 to get 0.
x\left(x-12\right)=0
Factor out x.
x=0 x=12
To find equation solutions, solve x=0 and x-12=0.
x^{2}-12x+36=36
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.
x^{2}-12x+36-36=0
Subtract 36 from both sides.
x^{2}-12x=0
Subtract 36 from 36 to get 0.
x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±12}{2}
Take the square root of \left(-12\right)^{2}.
x=\frac{12±12}{2}
The opposite of -12 is 12.
x=\frac{24}{2}
Now solve the equation x=\frac{12±12}{2} when ± is plus. Add 12 to 12.
x=12
Divide 24 by 2.
x=\frac{0}{2}
Now solve the equation x=\frac{12±12}{2} when ± is minus. Subtract 12 from 12.
x=0
Divide 0 by 2.
x=12 x=0
The equation is now solved.
\sqrt{\left(x-6\right)^{2}}=\sqrt{36}
Take the square root of both sides of the equation.
x-6=6 x-6=-6
Simplify.
x=12 x=0
Add 6 to both sides of the equation.
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