Solve X=(2x+3)^2quadB=

$$X = ( 2 x + 3 ) ^ { 2 } \quad B =$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B=\frac{X}{\left(2x+3\right)^{2}}\text{, }&x\neq -\frac{3}{2}\\B\in \mathrm{C}\text{, }&X=0\text{ and }x=-\frac{3}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$X=\left(4x^{2}+12x+9\right)B$$

Use the distributive property to multiply 4x^{2}+12x+9 by B.

$$X=4x^{2}B+12xB+9B$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}B+12xB+9B=X$$

Combine all terms containing B.

$$\left(4x^{2}+12x+9\right)B=X$$

Divide both sides by 4x^{2}+12x+9.

$$\frac{\left(4x^{2}+12x+9\right)B}{4x^{2}+12x+9}=\frac{X}{4x^{2}+12x+9}$$

Dividing by 4x^{2}+12x+9 undoes the multiplication by 4x^{2}+12x+9.

$$B=\frac{X}{4x^{2}+12x+9}$$

Divide X by 4x^{2}+12x+9.

$$B=\frac{X}{\left(2x+3\right)^{2}}$$

Solve for B

$$\left\{\begin{matrix}B=\frac{X}{\left(2x+3\right)^{2}}\text{, }&x\neq -\frac{3}{2}\\B\in \mathrm{R}\text{, }&X=0\text{ and }x=-\frac{3}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$X=\left(4x^{2}+12x+9\right)B$$

Use the distributive property to multiply 4x^{2}+12x+9 by B.

$$X=4x^{2}B+12xB+9B$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}B+12xB+9B=X$$

Combine all terms containing B.

$$\left(4x^{2}+12x+9\right)B=X$$

Divide both sides by 4x^{2}+12x+9.

$$\frac{\left(4x^{2}+12x+9\right)B}{4x^{2}+12x+9}=\frac{X}{4x^{2}+12x+9}$$

Dividing by 4x^{2}+12x+9 undoes the multiplication by 4x^{2}+12x+9.

$$B=\frac{X}{4x^{2}+12x+9}$$

Divide X by 4x^{2}+12x+9.

$$B=\frac{X}{\left(2x+3\right)^{2}}$$

Solve for X

$$X=B\left(2x+3\right)^{2}$$