15(t)+4.9*(t^2)=44

Solution for 15(t)+4.9*(t^2)=44 equation:

15(t)+4.9(t^2)=44

We move all terms to the left:

15(t)+4.9(t^2)-(44)=0

determiningTheFunctionDomain
4.9t^2+15t-44=0

a = 4.9; b = 15; c = -44;
Δ = b2-4ac
Δ = 152-4·4.9·(-44)
Δ = 1087.4
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{-(15)-\sqrt{1087.4}}{2*4.9}=\frac{-15-\sqrt{1087.4}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{1087.4}}{2*4.9}=\frac{-15+\sqrt{1087.4}}{9.8}
$