Answer:
1
Distribute
(
+
2
)
(
+
2
)
=
3
{\color{#c92786}{(x+2)(x+2)}}=3x
(
+
2
)
+
2
(
+
2
)
=
3
{\color{#c92786}{x(x+2)+2(x+2)}}=3x
2
Distribute
(
+
2
)
+
2
(
+
2
)
=
3
{\color{#c92786}{x(x+2)}}+2(x+2)=3x
2
+
2
+
2
(
+
2
)
=
3
{\color{#c92786}{x^{2}+2x}}+2(x+2)=3x
3
Distribute
2
+
2
+
2
(
+
2
)
=
3
x^{2}+2x+{\color{#c92786}{2(x+2)}}=3x
2
+
2
+
2
+
4
=
3
x^{2}+2x+{\color{#c92786}{2x+4}}=3x
4
Combine like terms
2
+
2
+
2
+
4
=
3
x^{2}+{\color{#c92786}{2x}}+{\color{#c92786}{2x}}+4=3x
2
+
4
+
4
=
3
x^{2}+{\color{#c92786}{4x}}+4=3x
5
Move terms to the left side
2
+
4
+
4
=
3
x^{2}+4x+4=3x
2
+
4
+
4
−
3
=
0
x^{2}+4x+4-3x=0
6
Combine like terms
2
+
4
+
4
−
3
=
0
x^{2}+{\color{#c92786}{4x}}+4{\color{#c92786}{-3x}}=0
2
+
1
+
4
=
0
x^{2}+{\color{#c92786}{1x}}+4=0
7
Multiply by 1
2
+
1
+
4
=
0
x^{2}+1x+4=0
2
+
+
4
=
0
x^{2}+x+4=0
8
Use the quadratic formula
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
2
+
+
4
=
0
x^{2}+x+4=0
=
1
a={\color{#c92786}{1}}
=
1
b={\color{#e8710a}{1}}
=
4
c={\color{#129eaf}{4}}
=
−
1
±
1
2
−
4
⋅
1
⋅
4
√
2
⋅
1
x=\frac{-{\color{#e8710a}{1}} \pm \sqrt{{\color{#e8710a}{1}}^{2}-4 \cdot {\color{#c92786}{1}} \cdot {\color{#129eaf}{4}}}}{2 \cdot {\color{#c92786}{1}}}
9
Simplify
Evaluate the exponent
Multiply the numbers
Subtract the numbers
Multiply the numbers
=
−
1
±
−
1
5
√
2
x=\frac{-1 \pm \sqrt{-15}}{2}
10
No real solutions because the discriminant is negative
The square root of a negative number is not a real number
=
−
1
5
Step-by-step explanation:
