4=-16t^2+120t+4

Solution for 4=-16t^2+120t+4 equation:

4=-16t^2+120t+4

We move all terms to the left:

4-(-16t^2+120t+4)=0

We get rid of parentheses

16t^2-120t-4+4=0

We add all the numbers together, and all the variables

16t^2-120t=0

a = 16; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·16·0
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*16}=\frac{240}{32}
=7+1/2
$