First we need to factor trinomial $ \color{blue}{ x^2-18x-144 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-18x-144 = 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:

$$ \color{blue}{ b = -18 } ~ \text{ and } ~ \color{red}{ c = -144 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -18 } $ and multiply to $ \color{red}{ -144 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -144 }$.

PRODUCT = -144
-1     144	1     -144
-2     72	2     -72
-3     48	3     -48
-4     36	4     -36
-6     24	6     -24
-8     18	8     -18
-9     16	9     -16
-12     12	12     -12

Step 3: Find out which pair sums up to $\color{blue}{ b = -18 }$

PRODUCT = -144 and SUM = -18
-1     144	1     -144
-2     72	2     -72
-3     48	3     -48
-4     36	4     -36
-6     24	6     -24
-8     18	8     -18
-9     16	9     -16
-12     12	12     -12

Step 4: Put 6 and -24 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-18x-144 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-18x-144 & = (x + 6)(x -24) \end{aligned} $$

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

$$ \begin{array}{ccc} \begin{aligned} x+6 &= 0 \\ x &= -6 \end{aligned} & ~ & \begin{aligned} x-24 &= 0 \\ x &= 24 \end{aligned} \end{array} $$

Quadratic Equation Solver