Solve 0=1/-(x+3)^2+5

$$0 = \frac { 1 } { - ( x + 3 ) ^ { 2 } } + 5$$

Solve for x

$$x=-\frac{\sqrt{5}}{5}-3\approx -3.447213595$$

$$x=\frac{\sqrt{5}}{5}-3\approx -2.552786405$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$0=\frac{1}{-\left(x^{2}+6x+9\right)}+5$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$0=\frac{1}{-x^{2}-6x-9}+5$$

Factor -x^{2}-6x-9.

$$0=\frac{1}{\left(-x-3\right)\left(x+3\right)}+5$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}.

$$0=\frac{1}{\left(-x-3\right)\left(x+3\right)}+\frac{5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}$$

Since \frac{1}{\left(-x-3\right)\left(x+3\right)} and \frac{5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$0=\frac{1+5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}$$

Do the multiplications in 1+5\left(-x-3\right)\left(x+3\right).

$$0=\frac{1-5x^{2}-15x-15x-45}{\left(-x-3\right)\left(x+3\right)}$$

Combine like terms in 1-5x^{2}-15x-15x-45.

$$0=\frac{-44-5x^{2}-30x}{\left(-x-3\right)\left(x+3\right)}$$

Use the distributive property to multiply -x-3 by x+3 and combine like terms.

$$0=\frac{-44-5x^{2}-30x}{-x^{2}-6x-9}$$

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

$$\frac{-44-5x^{2}-30x}{-x^{2}-6x-9}=0$$

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by \left(-x-3\right)\left(x+3\right).

$$-44-5x^{2}-30x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}-30x-44=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -30 for b, and -44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\left(-5\right)\left(-44\right)}}{2\left(-5\right)}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\left(-5\right)\left(-44\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-30\right)±\sqrt{900+20\left(-44\right)}}{2\left(-5\right)}$$

Multiply 20 times -44.

$$x=\frac{-\left(-30\right)±\sqrt{900-880}}{2\left(-5\right)}$$

Add 900 to -880.

$$x=\frac{-\left(-30\right)±\sqrt{20}}{2\left(-5\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-30\right)±2\sqrt{5}}{2\left(-5\right)}$$

The opposite of -30 is 30.

$$x=\frac{30±2\sqrt{5}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{30±2\sqrt{5}}{-10}$$

Now solve the equation x=\frac{30±2\sqrt{5}}{-10} when ± is plus. Add 30 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+30}{-10}$$

Divide 30+2\sqrt{5} by -10.

$$x=-\frac{\sqrt{5}}{5}-3$$

Now solve the equation x=\frac{30±2\sqrt{5}}{-10} when ± is minus. Subtract 2\sqrt{5} from 30.

$$x=\frac{30-2\sqrt{5}}{-10}$$

Divide 30-2\sqrt{5} by -10.

$$x=\frac{\sqrt{5}}{5}-3$$

The equation is now solved.

$$x=-\frac{\sqrt{5}}{5}-3 x=\frac{\sqrt{5}}{5}-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$0=\frac{1}{-\left(x^{2}+6x+9\right)}+5$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$0=\frac{1}{-x^{2}-6x-9}+5$$

Factor -x^{2}-6x-9.

$$0=\frac{1}{\left(-x-3\right)\left(x+3\right)}+5$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}.

$$0=\frac{1}{\left(-x-3\right)\left(x+3\right)}+\frac{5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}$$

Since \frac{1}{\left(-x-3\right)\left(x+3\right)} and \frac{5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$0=\frac{1+5\left(-x-3\right)\left(x+3\right)}{\left(-x-3\right)\left(x+3\right)}$$

Do the multiplications in 1+5\left(-x-3\right)\left(x+3\right).

$$0=\frac{1-5x^{2}-15x-15x-45}{\left(-x-3\right)\left(x+3\right)}$$

Combine like terms in 1-5x^{2}-15x-15x-45.

$$0=\frac{-44-5x^{2}-30x}{\left(-x-3\right)\left(x+3\right)}$$

Use the distributive property to multiply -x-3 by x+3 and combine like terms.

$$0=\frac{-44-5x^{2}-30x}{-x^{2}-6x-9}$$

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

$$\frac{-44-5x^{2}-30x}{-x^{2}-6x-9}=0$$

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by \left(-x-3\right)\left(x+3\right).

$$-44-5x^{2}-30x=0$$

Add 44 to both sides. Anything plus zero gives itself.

$$-5x^{2}-30x=44$$

Divide both sides by -5.

$$\frac{-5x^{2}-30x}{-5}=\frac{44}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{30}{-5}\right)x=\frac{44}{-5}$$

Divide -30 by -5.

$$x^{2}+6x=\frac{44}{-5}$$

Divide 44 by -5.

$$x^{2}+6x=-\frac{44}{5}$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-\frac{44}{5}+3^{2}$$

Square 3.

$$x^{2}+6x+9=-\frac{44}{5}+9$$

Add -\frac{44}{5} to 9.

$$x^{2}+6x+9=\frac{1}{5}$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=\frac{1}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{\frac{1}{5}}$$

Simplify.

$$x+3=\frac{\sqrt{5}}{5} x+3=-\frac{\sqrt{5}}{5}$$

Subtract 3 from both sides of the equation.

$$x=\frac{\sqrt{5}}{5}-3 x=-\frac{\sqrt{5}}{5}-3$$