(.05t^2)-10t+200=0

Solution for (.05t^2)-10t+200=0 equation:

(.05t^2)-10t+200=0

We add all the numbers together, and all the variables

-10t+(.05t^2)+200=0

We get rid of parentheses

.05t^2-10t+200=0

a = .05; b = -10; c = +200;
Δ = b2-4ac
Δ = -102-4·.05·200
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{15}}{2*.05}=\frac{10-2\sqrt{15}}{0.1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{15}}{2*.05}=\frac{10+2\sqrt{15}}{0.1}
$