To solve the equation, factor $x^{2}+5x-1800$ 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.
$$a+b=5$$ $$ab=-1800$$
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 $-1800$.
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-40\right)\left(x+45\right)$$
To find equation solutions, solve $x-40=0$ and $x+45=0$.
$$x=40$$ $$x=-45$$
Steps Using Factoring By Grouping
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}+5x-1800=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-1800$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=1\left(-1800\right)=-1800$$
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 $-1800$.
Rewrite $x^{2}+5x-1800$ as $\left(x^{2}-40x\right)+\left(45x-1800\right)$.
$$\left(x^{2}-40x\right)+\left(45x-1800\right)$$
Factor out $x$ in the first and $45$ in the second group.
$$x\left(x-40\right)+45\left(x-40\right)$$
Factor out common term $x-40$ by using distributive property.
$$\left(x-40\right)\left(x+45\right)$$
To find equation solutions, solve $x-40=0$ and $x+45=0$.
$$x=40$$ $$x=-45$$
Steps Using the Quadratic Formula
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}+5x-1800=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $5$ for $b$, and $-1800$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{-5±85}{2}$ when $±$ is plus. Add $-5$ to $85$.
$$x=\frac{80}{2}$$
Divide $80$ by $2$.
$$x=40$$
Now solve the equation $x=\frac{-5±85}{2}$ when $±$ is minus. Subtract $85$ from $-5$.
$$x=-\frac{90}{2}$$
Divide $-90$ by $2$.
$$x=-45$$
The equation is now solved.
$$x=40$$ $$x=-45$$
Steps for Completing the Square
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}+5x-1800=0$$
Add $1800$ to both sides. Anything plus zero gives itself.
$$x^{2}+5x=1800$$
Divide $5$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{2}$. Then add the square of $\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
