First we need to factor trinomial $ \color{blue}{ x^2-2x-15 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-2x-15 = 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:

$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = -15 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ -15 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -15 }$.

PRODUCT = -15
-1     15	1     -15
-3     5	3     -5

Step 3: Find out which pair sums up to $\color{blue}{ b = -2 }$

PRODUCT = -15 and SUM = -2
-1     15	1     -15
-3     5	3     -5

Step 4: Put 3 and -5 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-2x-15 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-2x-15 & = (x + 3)(x -5) \end{aligned} $$

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

$$ \begin{array}{ccc} \begin{aligned} x+3 &= 0 \\ x &= -3 \end{aligned} & ~ & \begin{aligned} x-5 &= 0 \\ x &= 5 \end{aligned} \end{array} $$

Quadratic Equation Solver