200-16*t^2-10*t=0

Solution for 200-16*t^2-10*t=0 equation:

200-16t^2-10t=0

a = -16; b = -10; c = +200;
Δ = b2-4ac
Δ = -102-4·(-16)·200
Δ = 12900
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:

$\sqrt{\Delta}=\sqrt{12900}=\sqrt{100*129}=\sqrt{100}*\sqrt{129}=10\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{129}}{2*-16}=\frac{10-10\sqrt{129}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{129}}{2*-16}=\frac{10+10\sqrt{129}}{-32}
$