50=10x+0.5x^2

Solution for 50=10x+0.5x^2 equation:

50=10x+0.5x^2

We move all terms to the left:

50-(10x+0.5x^2)=0

We get rid of parentheses

-0.5x^2-10x+50=0

a = -0.5; b = -10; c = +50;
Δ = b2-4ac
Δ = -102-4·(-0.5)·50
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{2}}{2*-0.5}=\frac{10-10\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{2}}{2*-0.5}=\frac{10+10\sqrt{2}}{-1}
$