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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2+9x+x+3)-120=0

We get rid of parentheses

3x^2+9x+x+3-120=0

We add all the numbers together, and all the variables

3x^2+10x-117=0

a = 3; b = 10; c = -117;
Δ = b2-4ac
Δ = 102-4·3·(-117)
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{94}}{2*3}=\frac{-10-4\sqrt{94}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{94}}{2*3}=\frac{-10+4\sqrt{94}}{6}
$