40=16t^2+80t+4

Solution for 40=16t^2+80t+4 equation:

40=16t^2+80t+4

We move all terms to the left:

40-(16t^2+80t+4)=0

We get rid of parentheses

-16t^2-80t-4+40=0

We add all the numbers together, and all the variables

-16t^2-80t+36=0

a = -16; b = -80; c = +36;
Δ = b2-4ac
Δ = -802-4·(-16)·36
Δ = 8704
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:

$\sqrt{\Delta}=\sqrt{8704}=\sqrt{256*34}=\sqrt{256}*\sqrt{34}=16\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{34}}{2*-16}=\frac{80-16\sqrt{34}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{34}}{2*-16}=\frac{80+16\sqrt{34}}{-32}
$