Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+10\right)^{2}$.
$$x^{2}+20x+100=36$$
Subtract $36$ from both sides.
$$x^{2}+20x+100-36=0$$
Subtract $36$ from $100$ to get $64$.
$$x^{2}+20x+64=0$$
To solve the equation, factor $x^{2}+20x+64$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=20$$ $$ab=64$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $64$.
$$1,64$$ $$2,32$$ $$4,16$$ $$8,8$$
Calculate the sum for each pair.
$$1+64=65$$ $$2+32=34$$ $$4+16=20$$ $$8+8=16$$
The solution is the pair that gives sum $20$.
$$a=4$$ $$b=16$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x+4\right)\left(x+16\right)$$
To find equation solutions, solve $x+4=0$ and $x+16=0$.
$$x=-4$$ $$x=-16$$
Steps Using Factoring By Grouping
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+10\right)^{2}$.
$$x^{2}+20x+100=36$$
Subtract $36$ from both sides.
$$x^{2}+20x+100-36=0$$
Subtract $36$ from $100$ to get $64$.
$$x^{2}+20x+64=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+64$. To find $a$ and $b$, set up a system to be solved.
$$a+b=20$$ $$ab=1\times 64=64$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $64$.
$$1,64$$ $$2,32$$ $$4,16$$ $$8,8$$
Calculate the sum for each pair.
$$1+64=65$$ $$2+32=34$$ $$4+16=20$$ $$8+8=16$$
The solution is the pair that gives sum $20$.
$$a=4$$ $$b=16$$
Rewrite $x^{2}+20x+64$ as $\left(x^{2}+4x\right)+\left(16x+64\right)$.
$$\left(x^{2}+4x\right)+\left(16x+64\right)$$
Factor out $x$ in the first and $16$ in the second group.
$$x\left(x+4\right)+16\left(x+4\right)$$
Factor out common term $x+4$ by using distributive property.
$$\left(x+4\right)\left(x+16\right)$$
To find equation solutions, solve $x+4=0$ and $x+16=0$.
$$x=-4$$ $$x=-16$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+10\right)^{2}$.
$$x^{2}+20x+100=36$$
Subtract $36$ from both sides.
$$x^{2}+20x+100-36=0$$
Subtract $36$ from $100$ to get $64$.
$$x^{2}+20x+64=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $20$ for $b$, and $64$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-20±\sqrt{20^{2}-4\times 64}}{2}$$
Square $20$.
$$x=\frac{-20±\sqrt{400-4\times 64}}{2}$$
Multiply $-4$ times $64$.
$$x=\frac{-20±\sqrt{400-256}}{2}$$
Add $400$ to $-256$.
$$x=\frac{-20±\sqrt{144}}{2}$$
Take the square root of $144$.
$$x=\frac{-20±12}{2}$$
Now solve the equation $x=\frac{-20±12}{2}$ when $±$ is plus. Add $-20$ to $12$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
Now solve the equation $x=\frac{-20±12}{2}$ when $±$ is minus. Subtract $12$ from $-20$.
$$x=-\frac{32}{2}$$
Divide $-32$ by $2$.
$$x=-16$$
The equation is now solved.
$$x=-4$$ $$x=-16$$
Steps for Completing the Square
Take the square root of both sides of the equation.
