500=50+4x+0.2x^2

Solution for 500=50+4x+0.2x^2 equation:

500=50+4x+0.2x^2

We move all terms to the left:

500-(50+4x+0.2x^2)=0

We get rid of parentheses

-0.2x^2-4x-50+500=0

We add all the numbers together, and all the variables

-0.2x^2-4x+450=0

a = -0.2; b = -4; c = +450;
Δ = b2-4ac
Δ = -42-4·(-0.2)·450
Δ = 376
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:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{94}}{2*-0.2}=\frac{4-2\sqrt{94}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{94}}{2*-0.2}=\frac{4+2\sqrt{94}}{-0.4}
$