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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-40t-t=0

We add all the numbers together, and all the variables

16t^2-41t=0

a = 16; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·16·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*16}=\frac{82}{32}
=2+9/16
$