-16t^2+35t+3=0

Solution for -16t^2+35t+3=0 equation:

-16t^2+35t+3=0

a = -16; b = 35; c = +3;
Δ = b2-4ac
Δ = 352-4·(-16)·3
Δ = 1417
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-\sqrt{1417}}{2*-16}=\frac{-35-\sqrt{1417}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{1417}}{2*-16}=\frac{-35+\sqrt{1417}}{-32}
$