-16t^2+44t+5=0

Solution for -16t^2+44t+5=0 equation:

-16t^2+44t+5=0

a = -16; b = 44; c = +5;
Δ = b2-4ac
Δ = 442-4·(-16)·5
Δ = 2256
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:

$\sqrt{\Delta}=\sqrt{2256}=\sqrt{16*141}=\sqrt{16}*\sqrt{141}=4\sqrt{141}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{141}}{2*-16}=\frac{-44-4\sqrt{141}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{141}}{2*-16}=\frac{-44+4\sqrt{141}}{-32}
$