(3y+2)(y-2)-(7y+3)(y-4)
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The quadratic equation is
3
y
2
−
7
y
+
4
=
0
The discriminant is
Δ
=
b
2
−
4
a
c
=
(
−
7
)
2
−
4
⋅
3
⋅
4
=
49
−
48
=
1
As
Δ
>
0
, there are
2
real roots
Therefore,
y
=
−
b
±
√
Δ
2
a
=
−
(
−
7
)
±
√
1
2
⋅
3
=
7
±
1
6
Therefore,
y
1
=
8
6
=
4
3
and
y
2
=
6
6
=
1
The solutions are
S
=
{
1
,
4
3
}
krish2112:
This is a question of class 8
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