-58=19t-4.9t^2

Solution for -58=19t-4.9t^2 equation:

-58=19t-4.9t^2

We move all terms to the left:

-58-(19t-4.9t^2)=0

We get rid of parentheses

4.9t^2-19t-58=0

a = 4.9; b = -19; c = -58;
Δ = b2-4ac
Δ = -192-4·4.9·(-58)
Δ = 1497.8
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{1497.8}}{2*4.9}=\frac{19-\sqrt{1497.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1497.8}}{2*4.9}=\frac{19+\sqrt{1497.8}}{9.8}
$