19=-16t^2+40t+3

Solution for 19=-16t^2+40t+3 equation:

19=-16t^2+40t+3

We move all terms to the left:

19-(-16t^2+40t+3)=0

We get rid of parentheses

16t^2-40t-3+19=0

We add all the numbers together, and all the variables

16t^2-40t+16=0

a = 16; b = -40; c = +16;
Δ = b2-4ac
Δ = -402-4·16·16
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24}{2*16}=\frac{64}{32}
=2
$