5=-16t^2+40t+0

Solution for 5=-16t^2+40t+0 equation:

5=-16t^2+40t+0

We move all terms to the left:

5-(-16t^2+40t+0)=0

We get rid of parentheses

16t^2-40t-0+5=0

We add all the numbers together, and all the variables

16t^2-40t+5=0

a = 16; b = -40; c = +5;
Δ = b2-4ac
Δ = -402-4·16·5
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{5}}{2*16}=\frac{40-16\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{5}}{2*16}=\frac{40+16\sqrt{5}}{32}
$