4.905t^2-6.88+0.6=0

Solution for 4.905t^2-6.88+0.6=0 equation:

4.905t^2-6.88+0.6=0

We add all the numbers together, and all the variables

4.905t^2-6.28=0

a = 4.905; b = 0; c = -6.28;
Δ = b2-4ac
Δ = 02-4·4.905·(-6.28)
Δ = 123.2136
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{123.2136}}{2*4.905}=\frac{0-\sqrt{123.2136}}{9.81}
=-\frac{\sqrt{}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{123.2136}}{2*4.905}=\frac{0+\sqrt{123.2136}}{9.81}
=\frac{\sqrt{}}{9.81}
$