3x+3x+3x^2+3x^2=240

Solution for 3x+3x+3x^2+3x^2=240 equation:

3x+3x+3x^2+3x^2=240

We move all terms to the left:

3x+3x+3x^2+3x^2-(240)=0

We add all the numbers together, and all the variables

6x^2+6x-240=0

a = 6; b = 6; c = -240;
Δ = b2-4ac
Δ = 62-4·6·(-240)
Δ = 5796
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:

$\sqrt{\Delta}=\sqrt{5796}=\sqrt{36*161}=\sqrt{36}*\sqrt{161}=6\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{161}}{2*6}=\frac{-6-6\sqrt{161}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{161}}{2*6}=\frac{-6+6\sqrt{161}}{12}
$