X~1.25. All you have to do is graph the function and see where the graph crosses the x-axis.
You subtract negative 4 and negative 12...think in a number line and you find negative 4 and then you decrease( which means you move to your left of the number line) and find negative 12 and then you count the spaces you took to get to negative 12
The value of y=29
the value of x=61
The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.
Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars
(not all zero) such that

which leads to the system of linear equations,

From the third equation, we have
, and substituting this into the second equation gives

and in turn,
. Substituting these into the first equation gives

which tells us that any value of
will work. If
, then
and
. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.
Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.