41=-16t^2+120t+4

Solution for 41=-16t^2+120t+4 equation:

41=-16t^2+120t+4

We move all terms to the left:

41-(-16t^2+120t+4)=0

We get rid of parentheses

16t^2-120t-4+41=0

We add all the numbers together, and all the variables

16t^2-120t+37=0

a = 16; b = -120; c = +37;
Δ = b2-4ac
Δ = -1202-4·16·37
Δ = 12032
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:

$\sqrt{\Delta}=\sqrt{12032}=\sqrt{256*47}=\sqrt{256}*\sqrt{47}=16\sqrt{47}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-16\sqrt{47}}{2*16}=\frac{120-16\sqrt{47}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+16\sqrt{47}}{2*16}=\frac{120+16\sqrt{47}}{32}
$