(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)-9x-18=0

We get rid of parentheses

2x^2+2x^2-8x+3x-9x-12-18=0

We add all the numbers together, and all the variables

4x^2-14x-30=0

a = 4; b = -14; c = -30;
Δ = b2-4ac
Δ = -142-4·4·(-30)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*4}=\frac{-12}{8}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*4}=\frac{40}{8}
=5
$