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

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

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

We multiply parentheses

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

We multiply parentheses ..

-(+2x^2-8x+3x-12)+4x+6=0

We get rid of parentheses

-2x^2+8x-3x+4x+12+6=0

We add all the numbers together, and all the variables

-2x^2+9x+18=0

a = -2; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·(-2)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*-2}=\frac{6}{-4}
=-1+1/2
$