16t^2-122t-99=0

Solution for 16t^2-122t-99=0 equation:

16t^2-122t-99=0

a = 16; b = -122; c = -99;
Δ = b2-4ac
Δ = -1222-4·16·(-99)
Δ = 21220
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:

$\sqrt{\Delta}=\sqrt{21220}=\sqrt{4*5305}=\sqrt{4}*\sqrt{5305}=2\sqrt{5305}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-122)-2\sqrt{5305}}{2*16}=\frac{122-2\sqrt{5305}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-122)+2\sqrt{5305}}{2*16}=\frac{122+2\sqrt{5305}}{32}
$