4.9t^2+4.6t-6.8=0

Solution for 4.9t^2+4.6t-6.8=0 equation:

4.9t^2+4.6t-6.8=0

a = 4.9; b = 4.6; c = -6.8;
Δ = b2-4ac
Δ = 4.62-4·4.9·(-6.8)
Δ = 154.44
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{-(4.6)-\sqrt{154.44}}{2*4.9}=\frac{-4.6-\sqrt{154.44}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.6)+\sqrt{154.44}}{2*4.9}=\frac{-4.6+\sqrt{154.44}}{9.8}
$