First we need to factor trinomial $ \color{blue}{ x^2+2x-3 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2+2x-3 = 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:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = -3 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ -3 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -3 }$.

PRODUCT = -3
-1     3	1     -3

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

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

Step 4: Put -1 and 3 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+2x-3 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+2x-3 & = (x -1)(x + 3) \end{aligned} $$

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

$$ \begin{array}{ccc} \begin{aligned} x-1 &= 0 \\ x &= 1 \end{aligned} & ~ & \begin{aligned} x+3 &= 0 \\ x &= -3 \end{aligned} \end{array} $$

Quadratic Equation Solver