0=-70-19t+4.9t^2

Solution for 0=-70-19t+4.9t^2 equation:

0=-70-19t+4.9t^2

We move all terms to the left:

0-(-70-19t+4.9t^2)=0

We add all the numbers together, and all the variables

-(-70-19t+4.9t^2)=0

We get rid of parentheses

-4.9t^2+19t+70=0

a = -4.9; b = 19; c = +70;
Δ = b2-4ac
Δ = 192-4·(-4.9)·70
Δ = 1733
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{1733}}{2*-4.9}=\frac{-19-\sqrt{1733}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1733}}{2*-4.9}=\frac{-19+\sqrt{1733}}{-9.8}
$