(2x-3)(x-6)+(2x-3)(x-4)=0

Solution for (2x-3)(x-6)+(2x-3)(x-4)=0 equation:

(2x-3)(x-6)+(2x-3)(x-4)=0

We multiply parentheses ..

(+2x^2-12x-3x+18)+(2x-3)(x-4)=0

We get rid of parentheses

2x^2-12x-3x+(2x-3)(x-4)+18=0

We multiply parentheses ..

2x^2+(+2x^2-8x-3x+12)-12x-3x+18=0

We add all the numbers together, and all the variables

2x^2+(+2x^2-8x-3x+12)-15x+18=0

We get rid of parentheses

2x^2+2x^2-8x-3x-15x+12+18=0

We add all the numbers together, and all the variables

4x^2-26x+30=0

a = 4; b = -26; c = +30;
Δ = b2-4ac
Δ = -262-4·4·30
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*4}=\frac{12}{8}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*4}=\frac{40}{8}
=5
$