50=0.5x^2-40x+600

Solution for 50=0.5x^2-40x+600 equation:

50=0.5x^2-40x+600

We move all terms to the left:

50-(0.5x^2-40x+600)=0

We get rid of parentheses

-0.5x^2+40x-600+50=0

We add all the numbers together, and all the variables

-0.5x^2+40x-550=0

a = -0.5; b = 40; c = -550;
Δ = b2-4ac
Δ = 402-4·(-0.5)·(-550)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{5}}{2*-0.5}=\frac{-40-10\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{5}}{2*-0.5}=\frac{-40+10\sqrt{5}}{-1}
$