4.9t^2-10.6t+.65=0

Solution for 4.9t^2-10.6t+.65=0 equation:

4.9t^2-10.6t+.65=0

We add all the numbers together, and all the variables

4.9t^2-10.6t+0.65=0

a = 4.9; b = -10.6; c = +0.65;
Δ = b2-4ac
Δ = -10.62-4·4.9·0.65
Δ = 99.62
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{-(-10.6)-\sqrt{99.62}}{2*4.9}=\frac{10.6-\sqrt{99.62}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10.6)+\sqrt{99.62}}{2*4.9}=\frac{10.6+\sqrt{99.62}}{9.8}
$