Solve x(3-2b)-1=x(2-3b-b^2)

$$x ( 3 - 2 b ) - 1 = x ( 2 - 3 b - b ^ { 2 } )$$

Solve for x (complex solution)

$$x=\frac{1}{b^{2}+b+1}$$

$$b\neq \frac{-1+\sqrt{3}i}{2}\text{ and }b\neq \frac{-\sqrt{3}i-1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by 3-2b.

$$3x-2xb-1=x\left(2-3b-b^{2}\right)$$

Use the distributive property to multiply x by 2-3b-b^{2}.

$$3x-2xb-1=2x-3xb-xb^{2}$$

Subtract 2x from both sides.

$$3x-2xb-1-2x=-3xb-xb^{2}$$

Combine 3x and -2x to get x.

$$x-2xb-1=-3xb-xb^{2}$$

Add 3xb to both sides.

$$x-2xb-1+3xb=-xb^{2}$$

Combine -2xb and 3xb to get xb.

$$x+xb-1=-xb^{2}$$

Add xb^{2} to both sides.

$$x+xb-1+xb^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x+xb+xb^{2}=1$$

Combine all terms containing x.

$$\left(1+b+b^{2}\right)x=1$$

The equation is in standard form.

$$\left(b^{2}+b+1\right)x=1$$

Divide both sides by 1+b+b^{2}.

$$\frac{\left(b^{2}+b+1\right)x}{b^{2}+b+1}=\frac{1}{b^{2}+b+1}$$

Dividing by 1+b+b^{2} undoes the multiplication by 1+b+b^{2}.

$$x=\frac{1}{b^{2}+b+1}$$

Solve for x

$$x=\frac{1}{b^{2}+b+1}$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by 3-2b.

$$3x-2xb-1=x\left(2-3b-b^{2}\right)$$

Use the distributive property to multiply x by 2-3b-b^{2}.

$$3x-2xb-1=2x-3xb-xb^{2}$$

Subtract 2x from both sides.

$$3x-2xb-1-2x=-3xb-xb^{2}$$

Combine 3x and -2x to get x.

$$x-2xb-1=-3xb-xb^{2}$$

Add 3xb to both sides.

$$x-2xb-1+3xb=-xb^{2}$$

Combine -2xb and 3xb to get xb.

$$x+xb-1=-xb^{2}$$

Add xb^{2} to both sides.

$$x+xb-1+xb^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x+xb+xb^{2}=1$$

Combine all terms containing x.

$$\left(1+b+b^{2}\right)x=1$$

The equation is in standard form.

$$\left(b^{2}+b+1\right)x=1$$

Divide both sides by 1+b+b^{2}.

$$\frac{\left(b^{2}+b+1\right)x}{b^{2}+b+1}=\frac{1}{b^{2}+b+1}$$

Dividing by 1+b+b^{2} undoes the multiplication by 1+b+b^{2}.

$$x=\frac{1}{b^{2}+b+1}$$

Solve for b (complex solution)

$$b=\frac{\sqrt{4x-3x^{2}}}{2x}-\frac{1}{2}$$

$$b=-\frac{\sqrt{4x-3x^{2}}}{2x}-\frac{1}{2}\text{, }x\neq 0$$

Solve for b

$$b=\frac{\sqrt{-3+\frac{4}{x}}-1}{2}$$

$$b=\frac{-\sqrt{-3+\frac{4}{x}}-1}{2}\text{, }x>0\text{ and }x\leq \frac{4}{3}$$