Solve left(x-6right)^2+8^2=100

$${ \left(x-6 \right) }^{ 2 } + { 8 }^{ 2 } = 100$$

Solve for x

$$x=0$$

$$x=12$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+8^{2}=100$$

Calculate 8 to the power of 2 and get 64.

$$x^{2}-12x+36+64=100$$

Add 36 and 64 to get 100.

$$x^{2}-12x+100=100$$

Subtract 100 from both sides.

$$x^{2}-12x+100-100=0$$

Subtract 100 from 100 to get 0.

$$x^{2}-12x=0$$

Factor out x.

$$x\left(x-12\right)=0$$

To find equation solutions, solve x=0 and x-12=0.

$$x=0 x=12$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+8^{2}=100$$

Calculate 8 to the power of 2 and get 64.

$$x^{2}-12x+36+64=100$$

Add 36 and 64 to get 100.

$$x^{2}-12x+100=100$$

Subtract 100 from both sides.

$$x^{2}-12x+100-100=0$$

Subtract 100 from 100 to get 0.

$$x^{2}-12x=0$$

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}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2}$$

Take the square root of \left(-12\right)^{2}.

$$x=\frac{-\left(-12\right)±12}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2}$$

Now solve the equation x=\frac{12±12}{2} when ± is plus. Add 12 to 12.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{12±12}{2} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=12 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+8^{2}=100$$

Calculate 8 to the power of 2 and get 64.

$$x^{2}-12x+36+64=100$$

Add 36 and 64 to get 100.

$$x^{2}-12x+100=100$$

Subtract 100 from both sides.

$$x^{2}-12x+100-100=0$$

Subtract 100 from 100 to get 0.

$$x^{2}-12x=0$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=36$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-6=6 x-6=-6$$

Add 6 to both sides of the equation.

$$x=12 x=0$$