100x^2+200x=800

Solution for 100x^2+200x=800 equation:

100x^2+200x=800

We move all terms to the left:

100x^2+200x-(800)=0

a = 100; b = 200; c = -800;
Δ = b2-4ac
Δ = 2002-4·100·(-800)
Δ = 360000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{360000}=600$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-600}{2*100}=\frac{-800}{200}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+600}{2*100}=\frac{400}{200}
=2
$