-16t^2+32t+40=0

Solution for -16t^2+32t+40=0 equation:

-16t^2+32t+40=0

a = -16; b = 32; c = +40;
Δ = b2-4ac
Δ = 322-4·(-16)·40
Δ = 3584
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:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{14}}{2*-16}=\frac{-32-16\sqrt{14}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{14}}{2*-16}=\frac{-32+16\sqrt{14}}{-32}
$