Solve for B (complex solution)
\left\{\begin{matrix}B=\frac{\left(3x+2\right)^{2}}{X}\text{, }&X\neq 0\\B\in \mathrm{C}\text{, }&x=-\frac{2}{3}\text{ and }X=0\end{matrix}\right.
Solve for X (complex solution)
\left\{\begin{matrix}X=\frac{\left(3x+2\right)^{2}}{B}\text{, }&B\neq 0\\X\in \mathrm{C}\text{, }&x=-\frac{2}{3}\text{ and }B=0\end{matrix}\right.
Solve for B
\left\{\begin{matrix}B=\frac{\left(3x+2\right)^{2}}{X}\text{, }&X\neq 0\\B\in \mathrm{R}\text{, }&x=-\frac{2}{3}\text{ and }X=0\end{matrix}\right.
Solve for X
\left\{\begin{matrix}X=\frac{\left(3x+2\right)^{2}}{B}\text{, }&B\neq 0\\X\in \mathrm{R}\text{, }&x=-\frac{2}{3}\text{ and }B=0\end{matrix}\right.
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XB=9x^{2}+12x+4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.
\frac{XB}{X}=\frac{\left(3x+2\right)^{2}}{X}
Divide both sides by X.
B=\frac{\left(3x+2\right)^{2}}{X}
Dividing by X undoes the multiplication by X.
XB=9x^{2}+12x+4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.
BX=9x^{2}+12x+4
The equation is in standard form.
\frac{BX}{B}=\frac{\left(3x+2\right)^{2}}{B}
Divide both sides by B.
X=\frac{\left(3x+2\right)^{2}}{B}
Dividing by B undoes the multiplication by B.
XB=9x^{2}+12x+4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.
\frac{XB}{X}=\frac{\left(3x+2\right)^{2}}{X}
Divide both sides by X.
B=\frac{\left(3x+2\right)^{2}}{X}
Dividing by X undoes the multiplication by X.
XB=9x^{2}+12x+4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.
BX=9x^{2}+12x+4
The equation is in standard form.
\frac{BX}{B}=\frac{\left(3x+2\right)^{2}}{B}
Divide both sides by B.
X=\frac{\left(3x+2\right)^{2}}{B}
Dividing by B undoes the multiplication by B.
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