Y=-16t^2+25t+5

Solution for Y=-16t^2+25t+5 equation:

=-16Y^2+25Y+5

We move all terms to the left:

-(-16Y^2+25Y+5)=0

We get rid of parentheses

16Y^2-25Y-5=0

a = 16; b = -25; c = -5;
Δ = b2-4ac
Δ = -252-4·16·(-5)
Δ = 945
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:

$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{105}}{2*16}=\frac{25-3\sqrt{105}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{105}}{2*16}=\frac{25+3\sqrt{105}}{32}
$