You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:

$$T_n=an^2+bn+c$$

You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.

This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.

Test cases

a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N 
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>

JavaScript (ES7), 48 bytes

Returns a Boolean value.

(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0

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How?

For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)

Case $aneq0$

The equation really is quadratic:

$$T_n=an^2+bn+c\
an^2+bn-d=0$$

And its discriminant is:

$$Delta=b^2+4ad$$

It admits an integer root if $Delta$ is non-negative and $sqrtDelta$ is an integer.

Case $a=0, bneq0$

The equation is linear:

$$T_n=bn+c\
bn=d\
n=fracdb$$

It admits an integer root if $dequiv0pmod b$.

Case $a=0, b=0$

The equation is not depending on $n$ anymore:

$$T_n=c\
d=0$$

Jelly,  11  10 bytes

_/Ær1Ẹ?%1Ạ

A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.

* admittedly taking a little liberty with "You may take input of these four numbers in any way".

Try it online! Or see a test-suite.

How?

_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
 / - reduce by:
_ - subtraction [c-T_n, b, a]
 ? - if...
 Ẹ - ...condition: any?
 Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
 1 - ...else: literal 1
 %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
 - note: (a+bi)%1 yields nan which is truthy
 Ạ - all? i.e. all had fractional parts?
 - note: all([]) yields 1

If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)

Jelly, 15 bytes

_3¦UÆr=Ḟ$;3ị=ɗẸ

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Built-in helps here but doesn’t handle a=b=0.

Wolfram Language (Mathematica), 38 bytes

Solve[n^2#+n#2+#3==#4,n,Integers]!=&

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