4.9=19t^2

Solution for 4.9=19t^2 equation:

4.9=19t^2

We move all terms to the left:

4.9-(19t^2)=0

a = -19; b = 0; c = +4.9;
Δ = b2-4ac
Δ = 02-4·(-19)·4.9
Δ = 372.4
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{-(0)-\sqrt{372.4}}{2*-19}=\frac{0-\sqrt{372.4}}{-38}
=-\frac{\sqrt{}}{-38}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{372.4}}{2*-19}=\frac{0+\sqrt{372.4}}{-38}
=\frac{\sqrt{}}{-38}
$