4.9t^2-9.1t-1.4=0

Solution for 4.9t^2-9.1t-1.4=0 equation:

4.9t^2-9.1t-1.4=0

a = 4.9; b = -9.1; c = -1.4;
Δ = b2-4ac
Δ = -9.12-4·4.9·(-1.4)
Δ = 110.25
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{-(-9.1)-\sqrt{110.25}}{2*4.9}=\frac{9.1-\sqrt{110.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9.1)+\sqrt{110.25}}{2*4.9}=\frac{9.1+\sqrt{110.25}}{9.8}
$