-16t^2-22t+108=0

Solution for -16t^2-22t+108=0 equation:

-16t^2-22t+108=0

a = -16; b = -22; c = +108;
Δ = b2-4ac
Δ = -222-4·(-16)·108
Δ = 7396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7396}=86$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-86}{2*-16}=\frac{-64}{-32}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+86}{2*-16}=\frac{108}{-32}
=-3+3/8
$