2.5=4.9t^2+19t

Solution for 2.5=4.9t^2+19t equation:

2.5=4.9t^2+19t

We move all terms to the left:

2.5-(4.9t^2+19t)=0

We get rid of parentheses

-4.9t^2-19t+2.5=0

a = -4.9; b = -19; c = +2.5;
Δ = b2-4ac
Δ = -192-4·(-4.9)·2.5
Δ = 410
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{410}}{2*-4.9}=\frac{19-\sqrt{410}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{410}}{2*-4.9}=\frac{19+\sqrt{410}}{-9.8}
$