48=(3x+2)(3x+6)

Solution for 48=(3x+2)(3x+6) equation:

48=(3x+2)(3x+6)

We move all terms to the left:

48-((3x+2)(3x+6))=0

We multiply parentheses ..

-((+9x^2+18x+6x+12))+48=0


We calculate terms in parentheses: -((+9x^2+18x+6x+12)), so:

(+9x^2+18x+6x+12)

We get rid of parentheses

9x^2+18x+6x+12

We add all the numbers together, and all the variables

9x^2+24x+12

Back to the equation:

-(9x^2+24x+12)


We get rid of parentheses

-9x^2-24x-12+48=0

We add all the numbers together, and all the variables

-9x^2-24x+36=0

a = -9; b = -24; c = +36;
Δ = b2-4ac
Δ = -242-4·(-9)·36
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{13}}{2*-9}=\frac{24-12\sqrt{13}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{13}}{2*-9}=\frac{24+12\sqrt{13}}{-18}
$