-16*t^2+14t+12=0

Solution for -16*t^2+14t+12=0 equation:

-16t^2+14t+12=0

a = -16; b = 14; c = +12;
Δ = b2-4ac
Δ = 142-4·(-16)·12
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{241}}{2*-16}=\frac{-14-2\sqrt{241}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{241}}{2*-16}=\frac{-14+2\sqrt{241}}{-32}
$