Solve 0=1/5left(x+5right)^2-1

$$0= \frac{ 1 }{ 5 } { \left(x+5 \right) }^{ 2 } -1$$

Solve for x

$$x=\sqrt{5}-5\approx -2.763932023$$

$$x=-\sqrt{5}-5\approx -7.236067977$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$0=\frac{1}{5}\left(x^{2}+10x+25\right)-1$$

Use the distributive property to multiply \frac{1}{5} by x^{2}+10x+25.

$$0=\frac{1}{5}x^{2}+2x+5-1$$

Subtract 1 from 5 to get 4.

$$0=\frac{1}{5}x^{2}+2x+4$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{5}x^{2}+2x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{5} for a, 2 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{5}\times 4}}{2\times \frac{1}{5}}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times \frac{1}{5}\times 4}}{2\times \frac{1}{5}}$$

Multiply -4 times \frac{1}{5}.

$$x=\frac{-2±\sqrt{4-\frac{4}{5}\times 4}}{2\times \frac{1}{5}}$$

Multiply -\frac{4}{5} times 4.

$$x=\frac{-2±\sqrt{4-\frac{16}{5}}}{2\times \frac{1}{5}}$$

Add 4 to -\frac{16}{5}.

$$x=\frac{-2±\sqrt{\frac{4}{5}}}{2\times \frac{1}{5}}$$

Take the square root of \frac{4}{5}.

$$x=\frac{-2±\frac{2\sqrt{5}}{5}}{2\times \frac{1}{5}}$$

Multiply 2 times \frac{1}{5}.

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

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

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

Divide -2+\frac{2\sqrt{5}}{5} by \frac{2}{5} by multiplying -2+\frac{2\sqrt{5}}{5} by the reciprocal of \frac{2}{5}.

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

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

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

Divide -2-\frac{2\sqrt{5}}{5} by \frac{2}{5} by multiplying -2-\frac{2\sqrt{5}}{5} by the reciprocal of \frac{2}{5}.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$0=\frac{1}{5}\left(x^{2}+10x+25\right)-1$$

Use the distributive property to multiply \frac{1}{5} by x^{2}+10x+25.

$$0=\frac{1}{5}x^{2}+2x+5-1$$

Subtract 1 from 5 to get 4.

$$0=\frac{1}{5}x^{2}+2x+4$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{5}x^{2}+2x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{5}x^{2}+2x=-4$$

Multiply both sides by 5.

$$\frac{\frac{1}{5}x^{2}+2x}{\frac{1}{5}}=-\frac{4}{\frac{1}{5}}$$

Dividing by \frac{1}{5} undoes the multiplication by \frac{1}{5}.

$$x^{2}+\frac{2}{\frac{1}{5}}x=-\frac{4}{\frac{1}{5}}$$

Divide 2 by \frac{1}{5} by multiplying 2 by the reciprocal of \frac{1}{5}.

$$x^{2}+10x=-\frac{4}{\frac{1}{5}}$$

Divide -4 by \frac{1}{5} by multiplying -4 by the reciprocal of \frac{1}{5}.

$$x^{2}+10x=-20$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-20+5^{2}$$

Square 5.

$$x^{2}+10x+25=-20+25$$

Add -20 to 25.

$$x^{2}+10x+25=5$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 5 from both sides of the equation.

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