-16t2+38t+5=0

Solution for -16t2+38t+5=0 equation:

-16t^2+38t+5=0

a = -16; b = 38; c = +5;
Δ = b2-4ac
Δ = 382-4·(-16)·5
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-42}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+42}{2*-16}=\frac{4}{-32}
=-1/8
$