40=40t-5t^2

Solution for 40=40t-5t^2 equation:

40=40t-5t^2

We move all terms to the left:

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

We get rid of parentheses

5t^2-40t+40=0

a = 5; b = -40; c = +40;
Δ = b2-4ac
Δ = -402-4·5·40
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{2}}{2*5}=\frac{40-20\sqrt{2}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{2}}{2*5}=\frac{40+20\sqrt{2}}{10}
$