16t^2-256t-200=0

Solution for 16t^2-256t-200=0 equation:

16t^2-256t-200=0

a = 16; b = -256; c = -200;
Δ = b2-4ac
Δ = -2562-4·16·(-200)
Δ = 78336
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:

$\sqrt{\Delta}=\sqrt{78336}=\sqrt{2304*34}=\sqrt{2304}*\sqrt{34}=48\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-256)-48\sqrt{34}}{2*16}=\frac{256-48\sqrt{34}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-256)+48\sqrt{34}}{2*16}=\frac{256+48\sqrt{34}}{32}
$