(x-3)(x+1)+(x-3)(2x-7)=0

Solution for (x-3)(x+1)+(x-3)(2x-7)=0 equation:

(x-3)(x+1)+(x-3)(2x-7)=0

We multiply parentheses ..

(+x^2+x-3x-3)+(x-3)(2x-7)=0

We get rid of parentheses

x^2+x-3x+(x-3)(2x-7)-3=0

We multiply parentheses ..

x^2+(+2x^2-7x-6x+21)+x-3x-3=0

We add all the numbers together, and all the variables

x^2+(+2x^2-7x-6x+21)-2x-3=0

We get rid of parentheses

x^2+2x^2-7x-6x-2x+21-3=0

We add all the numbers together, and all the variables

3x^2-15x+18=0

a = 3; b = -15; c = +18;
Δ = b2-4ac
Δ = -152-4·3·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3}{2*3}=\frac{12}{6}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3}{2*3}=\frac{18}{6}
=3
$