(2+4x)(3x-9)=18

Solution for (2+4x)(3x-9)=18 equation:

(2+4x)(3x-9)=18

We move all terms to the left:

(2+4x)(3x-9)-(18)=0

We add all the numbers together, and all the variables

(4x+2)(3x-9)-18=0

We multiply parentheses ..

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

We get rid of parentheses

12x^2-36x+6x-18-18=0

We add all the numbers together, and all the variables

12x^2-30x-36=0

a = 12; b = -30; c = -36;
Δ = b2-4ac
Δ = -302-4·12·(-36)
Δ = 2628
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2628}=\sqrt{36*73}=\sqrt{36}*\sqrt{73}=6\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{73}}{2*12}=\frac{30-6\sqrt{73}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{73}}{2*12}=\frac{30+6\sqrt{73}}{24}
$