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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-95t=0

a = 16; b = -95; c = 0;
Δ = b2-4ac
Δ = -952-4·16·0
Δ = 9025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9025}=95$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-95}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+95}{2*16}=\frac{190}{32}
=5+15/16
$