0=-16t^2+476t+120

Solution for 0=-16t^2+476t+120 equation:

0=-16t^2+476t+120

We move all terms to the left:

0-(-16t^2+476t+120)=0

We add all the numbers together, and all the variables

-(-16t^2+476t+120)=0

We get rid of parentheses

16t^2-476t-120=0

a = 16; b = -476; c = -120;
Δ = b2-4ac
Δ = -4762-4·16·(-120)
Δ = 234256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{234256}=484$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-476)-484}{2*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-476)+484}{2*16}=\frac{960}{32}
=30
$