Variable $x$ cannot be equal to any of the values $0,9$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-9\right)$, the least common multiple of $x,9x-x^{2},9-x$.
$$x-9-36-\left(-x\left(x-5\right)\right)=0$$
Subtract $36$ from $-9$ to get $-45$.
$$x-45-\left(-x\left(x-5\right)\right)=0$$
Use the distributive property to multiply $-x$ by $x-5$.
$$x-45-\left(-x^{2}+5x\right)=0$$
To find the opposite of $-x^{2}+5x$, find the opposite of each term.
$$x-45+x^{2}-5x=0$$
Combine $x$ and $-5x$ to get $-4x$.
$$-4x-45+x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-4x-45=0$$
To solve the equation, factor $x^{2}-4x-45$ 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.
$$a+b=-4$$ $$ab=-45$$
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. List all such integer pairs that give product $-45$.
$$1,-45$$ $$3,-15$$ $$5,-9$$
Calculate the sum for each pair.
$$1-45=-44$$ $$3-15=-12$$ $$5-9=-4$$
The solution is the pair that gives sum $-4$.
$$a=-9$$ $$b=5$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-9\right)\left(x+5\right)$$
To find equation solutions, solve $x-9=0$ and $x+5=0$.
$$x=9$$ $$x=-5$$
Variable $x$ cannot be equal to $9$.
$$x=-5$$
Steps Using Factoring By Grouping
Variable $x$ cannot be equal to any of the values $0,9$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-9\right)$, the least common multiple of $x,9x-x^{2},9-x$.
$$x-9-36-\left(-x\left(x-5\right)\right)=0$$
Subtract $36$ from $-9$ to get $-45$.
$$x-45-\left(-x\left(x-5\right)\right)=0$$
Use the distributive property to multiply $-x$ by $x-5$.
$$x-45-\left(-x^{2}+5x\right)=0$$
To find the opposite of $-x^{2}+5x$, find the opposite of each term.
$$x-45+x^{2}-5x=0$$
Combine $x$ and $-5x$ to get $-4x$.
$$-4x-45+x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-4x-45=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-45$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-4$$ $$ab=1\left(-45\right)=-45$$
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. List all such integer pairs that give product $-45$.
$$1,-45$$ $$3,-15$$ $$5,-9$$
Calculate the sum for each pair.
$$1-45=-44$$ $$3-15=-12$$ $$5-9=-4$$
The solution is the pair that gives sum $-4$.
$$a=-9$$ $$b=5$$
Rewrite $x^{2}-4x-45$ as $\left(x^{2}-9x\right)+\left(5x-45\right)$.
$$\left(x^{2}-9x\right)+\left(5x-45\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-9\right)+5\left(x-9\right)$$
Factor out common term $x-9$ by using distributive property.
$$\left(x-9\right)\left(x+5\right)$$
To find equation solutions, solve $x-9=0$ and $x+5=0$.
$$x=9$$ $$x=-5$$
Variable $x$ cannot be equal to $9$.
$$x=-5$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to any of the values $0,9$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-9\right)$, the least common multiple of $x,9x-x^{2},9-x$.
$$x-9-36-\left(-x\left(x-5\right)\right)=0$$
Subtract $36$ from $-9$ to get $-45$.
$$x-45-\left(-x\left(x-5\right)\right)=0$$
Use the distributive property to multiply $-x$ by $x-5$.
$$x-45-\left(-x^{2}+5x\right)=0$$
To find the opposite of $-x^{2}+5x$, find the opposite of each term.
$$x-45+x^{2}-5x=0$$
Combine $x$ and $-5x$ to get $-4x$.
$$-4x-45+x^{2}=0$$
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.
$$x^{2}-4x-45=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-4$ for $b$, and $-45$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{4±14}{2}$ when $±$ is plus. Add $4$ to $14$.
$$x=\frac{18}{2}$$
Divide $18$ by $2$.
$$x=9$$
Now solve the equation $x=\frac{4±14}{2}$ when $±$ is minus. Subtract $14$ from $4$.
$$x=-\frac{10}{2}$$
Divide $-10$ by $2$.
$$x=-5$$
The equation is now solved.
$$x=9$$ $$x=-5$$
Variable $x$ cannot be equal to $9$.
$$x=-5$$
Steps for Completing the Square
Variable $x$ cannot be equal to any of the values $0,9$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-9\right)$, the least common multiple of $x,9x-x^{2},9-x$.
$$x-9-36-\left(-x\left(x-5\right)\right)=0$$
Subtract $36$ from $-9$ to get $-45$.
$$x-45-\left(-x\left(x-5\right)\right)=0$$
Use the distributive property to multiply $-x$ by $x-5$.
$$x-45-\left(-x^{2}+5x\right)=0$$
To find the opposite of $-x^{2}+5x$, find the opposite of each term.
$$x-45+x^{2}-5x=0$$
Combine $x$ and $-5x$ to get $-4x$.
$$-4x-45+x^{2}=0$$
Add $45$ to both sides. Anything plus zero gives itself.
$$-4x+x^{2}=45$$
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 $x^{2}+bx=c$.
$$x^{2}-4x=45$$
Divide $-4$, the coefficient of the $x$ term, by $2$ to get $-2$. Then add the square of $-2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
