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Mathematics
Problem
(
t
+
1
)
2
+
c
=
0
(t+1)^2+c=0
(
t
+
1
)
2
+
c
=
0
In the equation above,
c
c
c
is a constant. The equation has solutions at
t
=
3
2
t=\frac{3}{2}
t
=
2
3
and
t
=
−
7
2
t=-\frac{7}{2}
t
=
−
2
7
. What is the value of
c
c
c
? A.
−
729
4
-\frac{729}{4}
−
4
729
B.
−
121
4
-\frac{121}{4}
−
4
121
C.
−
25
4
-\frac{25}{4}
−
4
25
D.
−
1
-1
−
1
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