3x+x+(3x)2=60

Solution for 3x+x+(3x)2=60 equation:

3x+x+(3x)2=60

We move all terms to the left:

3x+x+(3x)2-(60)=0

We add all the numbers together, and all the variables

3x^2+4x-60=0

a = 3; b = 4; c = -60;
Δ = b2-4ac
Δ = 42-4·3·(-60)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{46}}{2*3}=\frac{-4-4\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{46}}{2*3}=\frac{-4+4\sqrt{46}}{6}
$