0.5x^2+40x-300=300

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

0.5x^2+40x-300=300

We move all terms to the left:

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

We add all the numbers together, and all the variables

0.5x^2+40x-600=0

a = 0.5; b = 40; c = -600;
Δ = b2-4ac
Δ = 402-4·0.5·(-600)
Δ = 2800
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:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{7}}{2*0.5}=\frac{-40-20\sqrt{7}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{7}}{2*0.5}=\frac{-40+20\sqrt{7}}{1}
$