41=19.62t^2+25t-41

Solution for 41=19.62t^2+25t-41 equation:

41=19.62t^2+25t-41

We move all terms to the left:

41-(19.62t^2+25t-41)=0

We get rid of parentheses

-19.62t^2-25t+41+41=0

We add all the numbers together, and all the variables

-19.62t^2-25t+82=0

a = -19.62; b = -25; c = +82;
Δ = b2-4ac
Δ = -252-4·(-19.62)·82
Δ = 7060.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{7060.36}}{2*-19.62}=\frac{25-\sqrt{7060.36}}{-39.24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{7060.36}}{2*-19.62}=\frac{25+\sqrt{7060.36}}{-39.24}
$