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
$