(6x+12)(x-3)-(x+2)(2x+3)=0

Solution for (6x+12)(x-3)-(x+2)(2x+3)=0 equation:

(6x+12)(x-3)-(x+2)(2x+3)=0

We multiply parentheses ..

(+6x^2-18x+12x-36)-(x+2)(2x+3)=0

We get rid of parentheses

6x^2-18x+12x-(x+2)(2x+3)-36=0

We multiply parentheses ..

6x^2-(+2x^2+3x+4x+6)-18x+12x-36=0

We add all the numbers together, and all the variables

6x^2-(+2x^2+3x+4x+6)-6x-36=0

We get rid of parentheses

6x^2-2x^2-3x-4x-6x-6-36=0

We add all the numbers together, and all the variables

4x^2-13x-42=0

a = 4; b = -13; c = -42;
Δ = b2-4ac
Δ = -132-4·4·(-42)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-29}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+29}{2*4}=\frac{42}{8}
=5+1/4
$