(3x+3)(4x+1)=120

Solution for (3x+3)(4x+1)=120 equation:

(3x+3)(4x+1)=120

We move all terms to the left:

(3x+3)(4x+1)-(120)=0

We multiply parentheses ..

(+12x^2+3x+12x+3)-120=0

We get rid of parentheses

12x^2+3x+12x+3-120=0

We add all the numbers together, and all the variables

12x^2+15x-117=0

a = 12; b = 15; c = -117;
Δ = b2-4ac
Δ = 152-4·12·(-117)
Δ = 5841
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:

$\sqrt{\Delta}=\sqrt{5841}=\sqrt{9*649}=\sqrt{9}*\sqrt{649}=3\sqrt{649}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{649}}{2*12}=\frac{-15-3\sqrt{649}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{649}}{2*12}=\frac{-15+3\sqrt{649}}{24}
$