3=-16t^2+40t+4

Solution for 3=-16t^2+40t+4 equation:

3=-16t^2+40t+4

We move all terms to the left:

3-(-16t^2+40t+4)=0

We get rid of parentheses

16t^2-40t-4+3=0

We add all the numbers together, and all the variables

16t^2-40t-1=0

a = 16; b = -40; c = -1;
Δ = b2-4ac
Δ = -402-4·16·(-1)
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{26}}{2*16}=\frac{40-8\sqrt{26}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{26}}{2*16}=\frac{40+8\sqrt{26}}{32}
$