First we need to factor trinomial $ \color{blue}{ x^2-7x-18 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-7x-18 = 0} $.

Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -7 } ~ \text{ and } ~ \color{red}{ c = -18 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -7 } $ and multiply to $ \color{red}{ -18 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -18 }$.

PRODUCT = -18
-1     18	1     -18
-2     9	2     -9
-3     6	3     -6

Step 3: Find out which pair sums up to $\color{blue}{ b = -7 }$

PRODUCT = -18 and SUM = -7
-1     18	1     -18
-2     9	2     -9
-3     6	3     -6

Step 4: Put 2 and -9 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-7x-18 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-7x-18 & = (x + 2)(x -9) \end{aligned} $$

Step 5: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} x+2 &= 0 \\ x &= -2 \end{aligned} & ~ & \begin{aligned} x-9 &= 0 \\ x &= 9 \end{aligned} \end{array} $$

Quadratic Equation Solver