(-2x+3)(x+6)+(-2x+3)(4x-1)=0

Solution for (-2x+3)(x+6)+(-2x+3)(4x-1)=0 equation:

(-2x+3)(x+6)+(-2x+3)(4x-1)=0

We multiply parentheses ..

(-2x^2-12x+3x+18)+(-2x+3)(4x-1)=0

We get rid of parentheses

-2x^2-12x+3x+(-2x+3)(4x-1)+18=0

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-10x^2+5x+15=0

a = -10; b = 5; c = +15;
Δ = b2-4ac
Δ = 52-4·(-10)·15
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-25}{2*-10}=\frac{-30}{-20}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+25}{2*-10}=\frac{20}{-20}
=-1
$