2x^2+120x=600

Solution for 2x^2+120x=600 equation:

2x^2+120x=600

We move all terms to the left:

2x^2+120x-(600)=0

a = 2; b = 120; c = -600;
Δ = b2-4ac
Δ = 1202-4·2·(-600)
Δ = 19200
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:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-80\sqrt{3}}{2*2}=\frac{-120-80\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+80\sqrt{3}}{2*2}=\frac{-120+80\sqrt{3}}{4}
$