Variable $x$ cannot be equal to any of the values $-6,6$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+6\right)\left(x-6\right)^{2}$, the least common multiple of $x^{2}-12x+36,36-x^{2},x+6$.
Use the distributive property to multiply $6-x$ by $12$.
$$x+6+72-12x=\left(x-6\right)^{2}$$
Add $6$ and $72$ to get $78$.
$$x+78-12x=\left(x-6\right)^{2}$$
Combine $x$ and $-12x$ to get $-11x$.
$$-11x+78=\left(x-6\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$-11x+78=x^{2}-12x+36$$
Subtract $x^{2}$ from both sides.
$$-11x+78-x^{2}=-12x+36$$
Add $12x$ to both sides.
$$-11x+78-x^{2}+12x=36$$
Combine $-11x$ and $12x$ to get $x$.
$$x+78-x^{2}=36$$
Subtract $36$ from both sides.
$$x+78-x^{2}-36=0$$
Subtract $36$ from $78$ to get $42$.
$$x+42-x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+x+42=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+42$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=-42=-42$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-42$.
$$-1,42$$ $$-2,21$$ $$-3,14$$ $$-6,7$$
Calculate the sum for each pair.
$$-1+42=41$$ $$-2+21=19$$ $$-3+14=11$$ $$-6+7=1$$
The solution is the pair that gives sum $1$.
$$a=7$$ $$b=-6$$
Rewrite $-x^{2}+x+42$ as $\left(-x^{2}+7x\right)+\left(-6x+42\right)$.
$$\left(-x^{2}+7x\right)+\left(-6x+42\right)$$
Factor out $-x$ in the first and $-6$ in the second group.
$$-x\left(x-7\right)-6\left(x-7\right)$$
Factor out common term $x-7$ by using distributive property.
$$\left(x-7\right)\left(-x-6\right)$$
To find equation solutions, solve $x-7=0$ and $-x-6=0$.
$$x=7$$ $$x=-6$$
Variable $x$ cannot be equal to $-6$.
$$x=7$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to any of the values $-6,6$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+6\right)\left(x-6\right)^{2}$, the least common multiple of $x^{2}-12x+36,36-x^{2},x+6$.
Use the distributive property to multiply $6-x$ by $12$.
$$x+6+72-12x=\left(x-6\right)^{2}$$
Add $6$ and $72$ to get $78$.
$$x+78-12x=\left(x-6\right)^{2}$$
Combine $x$ and $-12x$ to get $-11x$.
$$-11x+78=\left(x-6\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$-11x+78=x^{2}-12x+36$$
Subtract $x^{2}$ from both sides.
$$-11x+78-x^{2}=-12x+36$$
Add $12x$ to both sides.
$$-11x+78-x^{2}+12x=36$$
Combine $-11x$ and $12x$ to get $x$.
$$x+78-x^{2}=36$$
Subtract $36$ from both sides.
$$x+78-x^{2}-36=0$$
Subtract $36$ from $78$ to get $42$.
$$x+42-x^{2}=0$$
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.
$$-x^{2}+x+42=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $1$ for $b$, and $42$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{-1±13}{-2}$ when $±$ is plus. Add $-1$ to $13$.
$$x=\frac{12}{-2}$$
Divide $12$ by $-2$.
$$x=-6$$
Now solve the equation $x=\frac{-1±13}{-2}$ when $±$ is minus. Subtract $13$ from $-1$.
$$x=-\frac{14}{-2}$$
Divide $-14$ by $-2$.
$$x=7$$
The equation is now solved.
$$x=-6$$ $$x=7$$
Variable $x$ cannot be equal to $-6$.
$$x=7$$
Steps for Completing the Square
Variable $x$ cannot be equal to any of the values $-6,6$ since division by zero is not defined. Multiply both sides of the equation by $\left(x+6\right)\left(x-6\right)^{2}$, the least common multiple of $x^{2}-12x+36,36-x^{2},x+6$.
Use the distributive property to multiply $6-x$ by $12$.
$$x+6+72-12x=\left(x-6\right)^{2}$$
Add $6$ and $72$ to get $78$.
$$x+78-12x=\left(x-6\right)^{2}$$
Combine $x$ and $-12x$ to get $-11x$.
$$-11x+78=\left(x-6\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-6\right)^{2}$.
$$-11x+78=x^{2}-12x+36$$
Subtract $x^{2}$ from both sides.
$$-11x+78-x^{2}=-12x+36$$
Add $12x$ to both sides.
$$-11x+78-x^{2}+12x=36$$
Combine $-11x$ and $12x$ to get $x$.
$$x+78-x^{2}=36$$
Subtract $78$ from both sides.
$$x-x^{2}=36-78$$
Subtract $78$ from $36$ to get $-42$.
$$x-x^{2}=-42$$
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$-x^{2}+x=-42$$
Divide both sides by $-1$.
$$\frac{-x^{2}+x}{-1}=-\frac{42}{-1}$$
Dividing by $-1$ undoes the multiplication by $-1$.
$$x^{2}+\frac{1}{-1}x=-\frac{42}{-1}$$
Divide $1$ by $-1$.
$$x^{2}-x=-\frac{42}{-1}$$
Divide $-42$ by $-1$.
$$x^{2}-x=42$$
Divide $-1$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{2}$. Then add the square of $-\frac{1}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
