-16t^2-22t+155=0

Solution for -16t^2-22t+155=0 equation:

-16t^2-22t+155=0

a = -16; b = -22; c = +155;
Δ = b2-4ac
Δ = -222-4·(-16)·155
Δ = 10404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10404}=102$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-102}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+102}{2*-16}=\frac{124}{-32}
=-3+7/8
$