2800=100(2x^2+x)

Solution for 2800=100(2x^2+x) equation:

2800=100(2x^2+x)

We move all terms to the left:

2800-(100(2x^2+x))=0


We calculate terms in parentheses: -(100(2x^2+x)), so:

100(2x^2+x)

We multiply parentheses

200x^2+100x

Back to the equation:

-(200x^2+100x)


We get rid of parentheses

-200x^2-100x+2800=0

a = -200; b = -100; c = +2800;
Δ = b2-4ac
Δ = -1002-4·(-200)·2800
Δ = 2250000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2250000}=1500$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-1500}{2*-200}=\frac{-1400}{-400}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+1500}{2*-200}=\frac{1600}{-400}
=-4
$