300=40x+300+0.5x^2-80x

Solution for 300=40x+300+0.5x^2-80x equation:

300=40x+300+0.5x^2-80x

We move all terms to the left:

300-(40x+300+0.5x^2-80x)=0

We get rid of parentheses

-0.5x^2-40x+80x-300+300=0

We add all the numbers together, and all the variables

-0.5x^2+40x=0

a = -0.5; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-0.5)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-0.5}=\frac{-80}{-1}
=+80
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-0.5}=\frac{0}{-1}
=0
$