-16t2+28t+3=0

Solution for -16t2+28t+3=0 equation:

-16t^2+28t+3=0

a = -16; b = 28; c = +3;
Δ = b2-4ac
Δ = 282-4·(-16)·3
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{61}}{2*-16}=\frac{-28-4\sqrt{61}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{61}}{2*-16}=\frac{-28+4\sqrt{61}}{-32}
$