3=-16t^2+40t+0

Solution for 3=-16t^2+40t+0 equation:

3=-16t^2+40t+0

We move all terms to the left:

3-(-16t^2+40t+0)=0

We get rid of parentheses

16t^2-40t-0+3=0

We add all the numbers together, and all the variables

16t^2-40t+3=0

a = 16; b = -40; c = +3;
Δ = b2-4ac
Δ = -402-4·16·3
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{22}}{2*16}=\frac{40-8\sqrt{22}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{22}}{2*16}=\frac{40+8\sqrt{22}}{32}
$