0=-4.9t^2+14.48t-5.3

Solution for 0=-4.9t^2+14.48t-5.3 equation:

0=-4.9t^2+14.48t-5.3

We move all terms to the left:

0-(-4.9t^2+14.48t-5.3)=0

We add all the numbers together, and all the variables

-(-4.9t^2+14.48t-5.3)=0

We get rid of parentheses

4.9t^2-14.48t+5.3=0

a = 4.9; b = -14.48; c = +5.3;
Δ = b2-4ac
Δ = -14.482-4·4.9·5.3
Δ = 105.7904
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{-(-14.48)-\sqrt{105.7904}}{2*4.9}=\frac{14.48-\sqrt{105.7904}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14.48)+\sqrt{105.7904}}{2*4.9}=\frac{14.48+\sqrt{105.7904}}{9.8}
$