Solve left(x-9right)^2=5^

$${ \left(x-9 \right) }^{ 2 } = { 5 }^{ }$$

Solve for x

$$x=\sqrt{5}+9\approx 11.236067977$$

$$x=9-\sqrt{5}\approx 6.763932023$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-9\right)^{2}.

$$x^{2}-18x+81=5^{1}$$

Calculate 5 to the power of 1 and get 5.

$$x^{2}-18x+81=5$$

Subtract 5 from both sides.

$$x^{2}-18x+81-5=0$$

Subtract 5 from 81 to get 76.

$$x^{2}-18x+76=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 76 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 76}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 76}}{2}$$

Multiply -4 times 76.

$$x=\frac{-\left(-18\right)±\sqrt{324-304}}{2}$$

Add 324 to -304.

$$x=\frac{-\left(-18\right)±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-\left(-18\right)±2\sqrt{5}}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{18±2\sqrt{5}}{2} when ± is plus. Add 18 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+18}{2}$$

Divide 18+2\sqrt{5} by 2.

$$x=\sqrt{5}+9$$

Now solve the equation x=\frac{18±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 18.

$$x=\frac{18-2\sqrt{5}}{2}$$

Divide 18-2\sqrt{5} by 2.

$$x=9-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+9 x=9-\sqrt{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-9\right)^{2}.

$$x^{2}-18x+81=5^{1}$$

Calculate 5 to the power of 1 and get 5.

$$x^{2}-18x+81=5$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-9=\sqrt{5} x-9=-\sqrt{5}$$

Add 9 to both sides of the equation.

$$x=\sqrt{5}+9 x=9-\sqrt{5}$$