-4.9t^2+44.1t+1=0

Solution for -4.9t^2+44.1t+1=0 equation:

-4.9t^2+44.1t+1=0

a = -4.9; b = 44.1; c = +1;
Δ = b2-4ac
Δ = 44.12-4·(-4.9)·1
Δ = 1964.41
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{-(44.1)-\sqrt{1964.41}}{2*-4.9}=\frac{-44.1-\sqrt{1964.41}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44.1)+\sqrt{1964.41}}{2*-4.9}=\frac{-44.1+\sqrt{1964.41}}{-9.8}
$