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

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

Solve for x

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

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

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

$$x^{2}+10x+25-5=0$$

Subtract 5 from 25 to get 20.

$$x^{2}+10x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 20}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-10±\sqrt{100-80}}{2}$$

Add 100 to -80.

$$x=\frac{-10±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-10±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{5}}{2} when ± is plus. Add -10 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-10}{2}$$

Divide -10+2\sqrt{5} by 2.

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

Now solve the equation x=\frac{-10±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -10.

$$x=\frac{-2\sqrt{5}-10}{2}$$

Divide -10-2\sqrt{5} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

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

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

$$x^{2}+10x+25-5=0$$

Subtract 5 from 25 to get 20.

$$x^{2}+10x+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

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

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.

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