-16t^2+38t+3=0

Solution for -16t^2+38t+3=0 equation:

-16t^2+38t+3=0

a = -16; b = 38; c = +3;
Δ = b2-4ac
Δ = 382-4·(-16)·3
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{409}}{2*-16}=\frac{-38-2\sqrt{409}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{409}}{2*-16}=\frac{-38+2\sqrt{409}}{-32}
$