(x-3)(x+1)+(x-3)(x+2)=0

Solution for (x-3)(x+1)+(x-3)(x+2)=0 equation:

(x-3)(x+1)+(x-3)(x+2)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

x^2+(+x^2+2x-3x-6)+x-3x-3=0

We add all the numbers together, and all the variables

x^2+(+x^2+2x-3x-6)-2x-3=0

We get rid of parentheses

x^2+x^2+2x-3x-2x-6-3=0

We add all the numbers together, and all the variables

2x^2-3x-9=0

a = 2; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·2·(-9)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*2}=\frac{12}{4}
=3
$