10=-4.9t^2+19t

Solution for 10=-4.9t^2+19t equation:

10=-4.9t^2+19t

We move all terms to the left:

10-(-4.9t^2+19t)=0

We get rid of parentheses

4.9t^2-19t+10=0

a = 4.9; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·4.9·10
Δ = 165
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{165}}{2*4.9}=\frac{19-\sqrt{165}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{165}}{2*4.9}=\frac{19+\sqrt{165}}{9.8}
$