Solve -(x+5)^2+36=0

$$- ( x + 5 ) ^ { 2 } + 36 = 0$$

Solve for x

$$x=1$$

$$x=-11$$

Steps Using Factoring By Grouping

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

$$-\left(x^{2}+10x+25\right)+36=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$-x^{2}-10x-25+36=0$$

Add -25 and 36 to get 11.

$$-x^{2}-10x+11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-11=-11$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-11$$

Rewrite -x^{2}-10x+11 as \left(-x^{2}+x\right)+\left(-11x+11\right).

$$\left(-x^{2}+x\right)+\left(-11x+11\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(-x+1\right)+11\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+11\right)$$

To find equation solutions, solve -x+1=0 and x+11=0.

$$x=1 x=-11$$

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}.

$$-\left(x^{2}+10x+25\right)+36=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$-x^{2}-10x-25+36=0$$

Add -25 and 36 to get 11.

$$-x^{2}-10x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\times 11}}{2\left(-1\right)}$$

Multiply 4 times 11.

$$x=\frac{-\left(-10\right)±\sqrt{100+44}}{2\left(-1\right)}$$

Add 100 to 44.

$$x=\frac{-\left(-10\right)±\sqrt{144}}{2\left(-1\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-10\right)±12}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±12}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±12}{-2}$$

Now solve the equation x=\frac{10±12}{-2} when ± is plus. Add 10 to 12.

$$x=\frac{22}{-2}$$

Divide 22 by -2.

$$x=-11$$

Now solve the equation x=\frac{10±12}{-2} when ± is minus. Subtract 12 from 10.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-11 x=1$$

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}.

$$-\left(x^{2}+10x+25\right)+36=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$-x^{2}-10x-25+36=0$$

Add -25 and 36 to get 11.

$$-x^{2}-10x+11=0$$

Subtract 11 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-10x=-11$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=-\frac{11}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=-\frac{11}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=-\frac{11}{-1}$$

Divide -11 by -1.

$$x^{2}+10x=11$$

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.

$$x^{2}+10x+5^{2}=11+5^{2}$$

Square 5.

$$x^{2}+10x+25=11+25$$

Add 11 to 25.

$$x^{2}+10x+25=36$$

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}.

$$\left(x+5\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+5=6 x+5=-6$$

Subtract 5 from both sides of the equation.

$$x=1 x=-11$$