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\[{a}^{2}-10.5a-25=0\]
+
−
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a^2-10.5a-25=0
a^2-10.5a-25=0
{a}^{2}-10.5a-25=0
1
Factor with quadratic formula.
1
In general, given
a
x
2
+
b
x
+
c
a{x}^{2}+bx+c
a
x
2
+
b
x
+
c
, the factored form is:
a
(
x
−
−
b
+
b
2
−
4
a
c
2
a
)
(
x
−
−
b
−
b
2
−
4
a
c
2
a
)
a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})
a
(
x
−
2
a
−
b
+
b
2
−
4
a
c
)
(
x
−
2
a
−
b
−
b
2
−
4
a
c
)
2
In this case,
a
=
1
a=1
a
=
1
,
b
=
−
10.5
b=-10.5
b
=
−
1
0
.
5
and
c
=
−
25
c=-25
c
=
−
2
5
.
(
a
−
10.5
+
(
−
10.5
)
2
−
4
×
−
25
2
)
(
a
−
10.5
−
(
−
10.5
)
2
−
4
×
−
25
2
)
(a-\frac{10.5+\sqrt{{(-10.5)}^{2}-4\times -25}}{2})(a-\frac{10.5-\sqrt{{(-10.5)}^{2}-4\times -25}}{2})
(
a
−
2
1
0
.
5
+
(
−
1
0
.
5
)
2
−
4
×
−
2
5
)
(
a
−
2
1
0
.
5
−
(
−
1
0
.
5
)
2
−
4
×
−
2
5
)
3
Simplify.
(
a
−
25
2
)
(
a
+
2
)
(a-\frac{25}{2})(a+2)
(
a
−
2
2
5
)
(
a
+
2
)
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!
(
a
−
25
2
)
(
a
+
2
)
=
0
(a-\frac{25}{2})(a+2)=0
(
a
−
2
2
5
)
(
a
+
2
)
=
0
2
Solve for
a
a
a
.
1
Ask: When will
(
a
−
25
2
)
(
a
+
2
)
(a-\frac{25}{2})(a+2)
(
a
−
2
2
5
)
(
a
+
2
)
equal zero?
When
a
−
25
2
=
0
a-\frac{25}{2}=0
a
−
2
2
5
=
0
or
a
+
2
=
0
a+2=0
a
+
2
=
0
2
Solve each of the 2 equations above.
a
=
25
2
,
−
2
a=\frac{25}{2},-2
a
=
2
2
5
,
−
2
To get access to all 'How?' and 'Why?' steps, join
Cymath Plus
!
a
=
25
2
,
−
2
a=\frac{25}{2},-2
a
=
2
2
5
,
−
2
Done
Decimal Form: 12.5, -2
a=25/2,-2
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