9
c
−
4
−
c
2
=
−
7
9
c
-
4
-
c
2
=
-
7
Move
7
7
to the left side of the equation by adding it to both sides.
9
c
−
4
−
c
2
+
7
=
0
9
c
-
4
-
c
2
+
7
=
0
Add
−
4
-
4
and
7
7
.
9
c
−
c
2
+
3
=
0
9
c
-
c
2
+
3
=
0
Factor
−
1
-
1
out of
9
c
−
c
2
+
3
9
c
-
c
2
+
3
.
Tap for more steps...
−
(
c
2
−
9
c
−
3
)
=
0
-
(
c
2
-
9
c
-
3
)
=
0
Multiply each term in
−
(
c
2
−
9
c
−
3
)
=
0
-
(
c
2
-
9
c
-
3
)
=
0
by
−
1
-
1
Tap for more steps...
c
2
−
9
c
−
3
=
0
c
2
-
9
c
-
3
=
0
Use the quadratic formula to find the solutions.
−
b
±
√
b
2
−
4
(
a
c
)
2
a
-
b
±
b
2
-
4
(
a
c
)
2
a
Substitute the values
a
=
1
a
=
1
,
b
=
−
9
b
=
-
9
, and
c
=
−
3
c
=
-
3
into the quadratic formula and solve for
c
c
.
9
±
√
(
−
9
)
2
−
4
⋅
(
1
⋅
−
3
)
2
⋅
1
9
±
(
-
9
)
2
-
4
⋅
(
1
⋅
-
3
)
2
⋅
1
Simplify.
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c
=
9
±
√
93
2
c
=
9
±
93
2
The final answer is the combination of both solutions.
c
=
9
+
√
93
2
,
9
−
√
93
2
c
=
9
+
93
2
,
9
-
93
2
The result can be shown in multiple forms.
Exact Form:
c
=
9
+
√
93
2
,
9
−
√
93
2
By making three groups, you can take all the information and find out how many people like each drink, there are 47 people in each group, so 47 people are accounted for. 50-47=3.
<span>I am almost down with advanced functions and I found it to be a little difficult. I've never been that strong in math so I expected this course to be hard for me. I tried pretty hard this year, and I think after the exam, I can maintain around a 85-87. I'm just wondering if calculus is a lot more difficult than functions to the people that have already taken both courses. I've heard multiple times that calculus can be easier, but I'm not quite sure.</span>