3x^2+48x=180

Solution for 3x^2+48x=180 equation:

3x^2+48x=180

We move all terms to the left:

3x^2+48x-(180)=0

a = 3; b = 48; c = -180;
Δ = b2-4ac
Δ = 482-4·3·(-180)
Δ = 4464
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:

$\sqrt{\Delta}=\sqrt{4464}=\sqrt{144*31}=\sqrt{144}*\sqrt{31}=12\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{31}}{2*3}=\frac{-48-12\sqrt{31}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{31}}{2*3}=\frac{-48+12\sqrt{31}}{6}
$