195=16t^2+120t

Solution for 195=16t^2+120t equation:

195=16t^2+120t

We move all terms to the left:

195-(16t^2+120t)=0

We get rid of parentheses

-16t^2-120t+195=0

a = -16; b = -120; c = +195;
Δ = b2-4ac
Δ = -1202-4·(-16)·195
Δ = 26880
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:

$\sqrt{\Delta}=\sqrt{26880}=\sqrt{256*105}=\sqrt{256}*\sqrt{105}=16\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-16\sqrt{105}}{2*-16}=\frac{120-16\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+16\sqrt{105}}{2*-16}=\frac{120+16\sqrt{105}}{-32}
$