2x^2+20x=600

Solution for 2x^2+20x=600 equation:

2x^2+20x=600

We move all terms to the left:

2x^2+20x-(600)=0

a = 2; b = 20; c = -600;
Δ = b2-4ac
Δ = 202-4·2·(-600)
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{13}}{2*2}=\frac{-20-20\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{13}}{2*2}=\frac{-20+20\sqrt{13}}{4}
$