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x^{2}+4x+4=3x+6
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.
x^{2}+4x+4-3x=6
Subtract 3x from both sides.
x^{2}+x+4=6
Combine 4x and -3x to get x.
x^{2}+x+4-6=0
Subtract 6 from both sides.
x^{2}+x-2=0
Subtract 6 from 4 to get -2.
a+b=1 ab=-2
To solve the equation, factor x^{2}+x-2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
a=-1 b=2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(x-1\right)\left(x+2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=1 x=-2
To find equation solutions, solve x-1=0 and x+2=0.
x^{2}+4x+4=3x+6
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.
x^{2}+4x+4-3x=6
Subtract 3x from both sides.
x^{2}+x+4=6
Combine 4x and -3x to get x.
x^{2}+x+4-6=0
Subtract 6 from both sides.
x^{2}+x-2=0
Subtract 6 from 4 to get -2.
a+b=1 ab=1\left(-2\right)=-2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
a=-1 b=2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(x^{2}-x\right)+\left(2x-2\right)
Rewrite x^{2}+x-2 as \left(x^{2}-x\right)+\left(2x-2\right).
x\left(x-1\right)+2\left(x-1\right)
Factor out x in the first and 2 in the second group.
\left(x-1\right)\left(x+2\right)
Factor out common term x-1 by using distributive property.
x=1 x=-2
To find equation solutions, solve x-1=0 and x+2=0.
x^{2}+4x+4=3x+6
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.
x^{2}+4x+4-3x=6
Subtract 3x from both sides.
x^{2}+x+4=6
Combine 4x and -3x to get x.
x^{2}+x+4-6=0
Subtract 6 from both sides.
x^{2}+x-2=0
Subtract 6 from 4 to get -2.
x=\frac{-1±\sqrt{1^{2}-4\left(-2\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-2\right)}}{2}
Square 1.
x=\frac{-1±\sqrt{1+8}}{2}
Multiply -4 times -2.
x=\frac{-1±\sqrt{9}}{2}
Add 1 to 8.
x=\frac{-1±3}{2}
Take the square root of 9.
x=\frac{2}{2}
Now solve the equation x=\frac{-1±3}{2} when ± is plus. Add -1 to 3.
x=1
Divide 2 by 2.
x=-\frac{4}{2}
Now solve the equation x=\frac{-1±3}{2} when ± is minus. Subtract 3 from -1.
x=-2
Divide -4 by 2.
x=1 x=-2
The equation is now solved.
x^{2}+4x+4=3x+6
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.
x^{2}+4x+4-3x=6
Subtract 3x from both sides.
x^{2}+x+4=6
Combine 4x and -3x to get x.
x^{2}+x=6-4
Subtract 4 from both sides.
x^{2}+x=2
Subtract 4 from 6 to get 2.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=2+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=2+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{9}{4}
Add 2 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}
Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{3}{2} x+\frac{1}{2}=-\frac{3}{2}
Simplify.
x=1 x=-2
Subtract \frac{1}{2} from both sides of the equation.