4.9t^2+5.5t-24=0

Solution for 4.9t^2+5.5t-24=0 equation:

4.9t^2+5.5t-24=0

a = 4.9; b = 5.5; c = -24;
Δ = b2-4ac
Δ = 5.52-4·4.9·(-24)
Δ = 500.65
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{-(5.5)-\sqrt{500.65}}{2*4.9}=\frac{-5.5-\sqrt{500.65}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.5)+\sqrt{500.65}}{2*4.9}=\frac{-5.5+\sqrt{500.65}}{9.8}
$