Factor out the common term \(9\).
\[9(4{x}^{2}+4x+1)=0\]
Rewrite \(4{x}^{2}+4x+1\) in the form \({a}^{2}+2ab+{b}^{2}\), where \(a=2x\) and \(b=1\).
\[9({(2x)}^{2}+2(2x)(1)+{1}^{2})=0\]
Use Square of Sum: \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\).
\[9{(2x+1)}^{2}=0\]
Divide both sides by \(9\).
\[{(2x+1)}^{2}=0\]
Take the square root of both sides.
\[2x+1=0\]
Subtract \(1\) from both sides.
\[2x=-1\]
Divide both sides by \(2\).
\[x=-\frac{1}{2}\]
Decimal Form: -0.5
x=-1/2
