2x^2+40x=200

Solution for 2x^2+40x=200 equation:

2x^2+40x=200

We move all terms to the left:

2x^2+40x-(200)=0

a = 2; b = 40; c = -200;
Δ = b2-4ac
Δ = 402-4·2·(-200)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{2}}{2*2}=\frac{-40-40\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{2}}{2*2}=\frac{-40+40\sqrt{2}}{4}
$