-16t^2-256t+300=0

Solution for -16t^2-256t+300=0 equation:

-16t^2-256t+300=0

a = -16; b = -256; c = +300;
Δ = b2-4ac
Δ = -2562-4·(-16)·300
Δ = 84736
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:

$\sqrt{\Delta}=\sqrt{84736}=\sqrt{256*331}=\sqrt{256}*\sqrt{331}=16\sqrt{331}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-256)-16\sqrt{331}}{2*-16}=\frac{256-16\sqrt{331}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-256)+16\sqrt{331}}{2*-16}=\frac{256+16\sqrt{331}}{-32}
$