200+20t=0,5t^2

Solution for 200+20t=0,5t^2 equation:

200+20t=0.5t^2

We move all terms to the left:

200+20t-(0.5t^2)=0

We get rid of parentheses

-0.5t^2+20t+200=0

a = -0.5; b = 20; c = +200;
Δ = b2-4ac
Δ = 202-4·(-0.5)·200
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*-0.5}=\frac{-20-20\sqrt{2}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*-0.5}=\frac{-20+20\sqrt{2}}{-1}
$