-16t^2+40t+25=0

Solution for -16t^2+40t+25=0 equation:

-16t^2+40t+25=0

a = -16; b = 40; c = +25;
Δ = b2-4ac
Δ = 402-4·(-16)·25
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{2}}{2*-16}=\frac{-40-40\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{2}}{2*-16}=\frac{-40+40\sqrt{2}}{-32}
$