2+14t+-4.9t2=0

Solution for 2+14t+-4.9t2=0 equation:

2+14t+-4.9t^2=0

We add all the numbers together, and all the variables

-4.9t^2+14t=0

a = -4.9; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-4.9)·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-4.9}=\frac{-28}{-9.8}
=2+6/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-4.9}=\frac{0}{-9.8}
=0
$