Y=-16t^2+48t+5

Solution for Y=-16t^2+48t+5 equation:

=-16Y^2+48Y+5

We move all terms to the left:

-(-16Y^2+48Y+5)=0

We get rid of parentheses

16Y^2-48Y-5=0

a = 16; b = -48; c = -5;
Δ = b2-4ac
Δ = -482-4·16·(-5)
Δ = 2624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{41}}{2*16}=\frac{48-8\sqrt{41}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{41}}{2*16}=\frac{48+8\sqrt{41}}{32}
$