5=-16t^2+160t

Solution for 5=-16t^2+160t equation:

5=-16t^2+160t

We move all terms to the left:

5-(-16t^2+160t)=0

We get rid of parentheses

16t^2-160t+5=0

a = 16; b = -160; c = +5;
Δ = b2-4ac
Δ = -1602-4·16·5
Δ = 25280
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:

$\sqrt{\Delta}=\sqrt{25280}=\sqrt{64*395}=\sqrt{64}*\sqrt{395}=8\sqrt{395}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{395}}{2*16}=\frac{160-8\sqrt{395}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{395}}{2*16}=\frac{160+8\sqrt{395}}{32}
$