-50=19.3t-4.9t^2

Solution for -50=19.3t-4.9t^2 equation:

-50=19.3t-4.9t^2

We move all terms to the left:

-50-(19.3t-4.9t^2)=0

We get rid of parentheses

4.9t^2-19.3t-50=0

a = 4.9; b = -19.3; c = -50;
Δ = b2-4ac
Δ = -19.32-4·4.9·(-50)
Δ = 1352.49
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{-(-19.3)-\sqrt{1352.49}}{2*4.9}=\frac{19.3-\sqrt{1352.49}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.3)+\sqrt{1352.49}}{2*4.9}=\frac{19.3+\sqrt{1352.49}}{9.8}
$