10=-19t+4.905t^2

Solution for 10=-19t+4.905t^2 equation:

10=-19t+4.905t^2

We move all terms to the left:

10-(-19t+4.905t^2)=0

We get rid of parentheses

-4.905t^2+19t+10=0

a = -4.905; b = 19; c = +10;
Δ = b2-4ac
Δ = 192-4·(-4.905)·10
Δ = 557.2
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{-(19)-\sqrt{557.2}}{2*-4.905}=\frac{-19-\sqrt{557.2}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{557.2}}{2*-4.905}=\frac{-19+\sqrt{557.2}}{-9.81}
$