(-16t^2)+100t+25=0

Solution for (-16t^2)+100t+25=0 equation:

(-16t^2)+100t+25=0

We get rid of parentheses

-16t^2+100t+25=0

a = -16; b = 100; c = +25;
Δ = b2-4ac
Δ = 1002-4·(-16)·25
Δ = 11600
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:

$\sqrt{\Delta}=\sqrt{11600}=\sqrt{400*29}=\sqrt{400}*\sqrt{29}=20\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{29}}{2*-16}=\frac{-100-20\sqrt{29}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{29}}{2*-16}=\frac{-100+20\sqrt{29}}{-32}
$