-4t^2+8t+25=0

Solution for -4t^2+8t+25=0 equation:

-4t^2+8t+25=0

a = -4; b = 8; c = +25;
Δ = b2-4ac
Δ = 82-4·(-4)·25
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{29}}{2*-4}=\frac{-8-4\sqrt{29}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{29}}{2*-4}=\frac{-8+4\sqrt{29}}{-8}
$