1 Use the Quadratic Formula.
1 In general, given
a
x
2
+
b
x
+
c
=
0
ax
2
+bx+c=0, there exists two solutions where:
x
=
−
b
+
b
2
−
4
a
c
2
a
,
−
b
−
b
2
−
4
a
c
2
a
x=
2a
−b+
b
2
−4ac
,
2a
−b−
b
2
−4ac
2 In this case,
a
=
3
a=3,
b
=
−
1
b=−1 and
c
=
9
c=9.
x
=
1
+
(
−
1
)
2
−
4
×
3
×
9
2
×
3
,
1
−
(
−
1
)
2
−
4
×
3
×
9
2
×
3
x
=
2×3
1+
(−1)
2
−4×3×9
,
2×3
1−
(−1)
2
−4×3×9
3 Simplify.
x
=
1
+
107
ı
6
,
1
−
107
ı
6
x=
6
1+
107
,
6
1−
107
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x
=
1
+
107
ı
6
,
1
−
107
ı
6
x=
6
1+
107
,
6
1−
107