-16t^2+16t+360=0

Solution for -16t^2+16t+360=0 equation:

-16t^2+16t+360=0

a = -16; b = 16; c = +360;
Δ = b2-4ac
Δ = 162-4·(-16)·360
Δ = 23296
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:

$\sqrt{\Delta}=\sqrt{23296}=\sqrt{256*91}=\sqrt{256}*\sqrt{91}=16\sqrt{91}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{91}}{2*-16}=\frac{-16-16\sqrt{91}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{91}}{2*-16}=\frac{-16+16\sqrt{91}}{-32}
$