40=-16t^2+50t+4

Solution for 40=-16t^2+50t+4 equation:

40=-16t^2+50t+4

We move all terms to the left:

40-(-16t^2+50t+4)=0

We get rid of parentheses

16t^2-50t-4+40=0

We add all the numbers together, and all the variables

16t^2-50t+36=0

a = 16; b = -50; c = +36;
Δ = b2-4ac
Δ = -502-4·16·36
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-14}{2*16}=\frac{36}{32}
=1+1/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+14}{2*16}=\frac{64}{32}
=2
$