0=16t^2+95t-120

Solution for 0=16t^2+95t-120 equation:

0=16t^2+95t-120

We move all terms to the left:

0-(16t^2+95t-120)=0

We add all the numbers together, and all the variables

-(16t^2+95t-120)=0

We get rid of parentheses

-16t^2-95t+120=0

a = -16; b = -95; c = +120;
Δ = b2-4ac
Δ = -952-4·(-16)·120
Δ = 16705
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-\sqrt{16705}}{2*-16}=\frac{95-\sqrt{16705}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{16705}}{2*-16}=\frac{95+\sqrt{16705}}{-32}
$