0=m^2+2m-3

Solution for 0=m^2+2m-3 equation:

0=m^2+2m-3

We move all terms to the left:

0-(m^2+2m-3)=0

We add all the numbers together, and all the variables

-(m^2+2m-3)=0

We get rid of parentheses

-m^2-2m+3=0

We add all the numbers together, and all the variables

-1m^2-2m+3=0

a = -1; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-1}=\frac{-2}{-2}
=1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-1}=\frac{6}{-2}
=-3
$