Solve (2x+3)(2x+3+b)-c=0

$$( 2 x + 3 ) ( 2 x + 3 + b ) - c = 0$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{4x^{2}+12x-c+9}{2x+3}\text{, }&x\neq -\frac{3}{2}\\b\in \mathrm{C}\text{, }&c=0\text{ and }x=-\frac{3}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x+3 by 2x+3+b and combine like terms.

$$4x^{2}+12x+2xb+9+3b-c=0$$

Subtract 4x^{2} from both sides. Anything subtracted from zero gives its negation.

$$12x+2xb+9+3b-c=-4x^{2}$$

Subtract 12x from both sides.

$$2xb+9+3b-c=-4x^{2}-12x$$

Subtract 9 from both sides.

$$2xb+3b-c=-4x^{2}-12x-9$$

Add c to both sides.

$$2xb+3b=-4x^{2}-12x-9+c$$

Combine all terms containing b.

$$\left(2x+3\right)b=-4x^{2}-12x-9+c$$

The equation is in standard form.

$$\left(2x+3\right)b=-4x^{2}-12x+c-9$$

Divide both sides by 2x+3.

$$\frac{\left(2x+3\right)b}{2x+3}=\frac{-\left(2x+3\right)^{2}+c}{2x+3}$$

Dividing by 2x+3 undoes the multiplication by 2x+3.

$$b=\frac{-\left(2x+3\right)^{2}+c}{2x+3}$$

Divide c-\left(2x+3\right)^{2} by 2x+3.

$$b=\frac{-4x^{2}-12x+c-9}{2x+3}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{4x^{2}+12x-c+9}{2x+3}\text{, }&x\neq -\frac{3}{2}\\b\in \mathrm{R}\text{, }&c=0\text{ and }x=-\frac{3}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x+3 by 2x+3+b and combine like terms.

$$4x^{2}+12x+2xb+9+3b-c=0$$

Subtract 4x^{2} from both sides. Anything subtracted from zero gives its negation.

$$12x+2xb+9+3b-c=-4x^{2}$$

Subtract 12x from both sides.

$$2xb+9+3b-c=-4x^{2}-12x$$

Subtract 9 from both sides.

$$2xb+3b-c=-4x^{2}-12x-9$$

Add c to both sides.

$$2xb+3b=-4x^{2}-12x-9+c$$

Combine all terms containing b.

$$\left(2x+3\right)b=-4x^{2}-12x-9+c$$

The equation is in standard form.

$$\left(2x+3\right)b=-4x^{2}-12x+c-9$$

Divide both sides by 2x+3.

$$\frac{\left(2x+3\right)b}{2x+3}=\frac{-\left(2x+3\right)^{2}+c}{2x+3}$$

Dividing by 2x+3 undoes the multiplication by 2x+3.

$$b=\frac{-\left(2x+3\right)^{2}+c}{2x+3}$$

Divide c-\left(2x+3\right)^{2} by 2x+3.

$$b=\frac{-4x^{2}-12x+c-9}{2x+3}$$

Solve for c

$$c=\left(2x+3\right)\left(2x+b+3\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x+3 by 2x+3+b and combine like terms.

$$4x^{2}+12x+2xb+9+3b-c=0$$

Subtract 4x^{2} from both sides. Anything subtracted from zero gives its negation.

$$12x+2xb+9+3b-c=-4x^{2}$$

Subtract 12x from both sides.

$$2xb+9+3b-c=-4x^{2}-12x$$

Subtract 2xb from both sides.

$$9+3b-c=-4x^{2}-12x-2xb$$

Subtract 9 from both sides.

$$3b-c=-4x^{2}-12x-2xb-9$$

Subtract 3b from both sides.

$$-c=-4x^{2}-12x-2xb-9-3b$$

The equation is in standard form.

$$-c=-4x^{2}-2bx-12x-3b-9$$

Divide both sides by -1.

$$\frac{-c}{-1}=-\frac{\left(2x+3\right)\left(2x+b+3\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$c=-\frac{\left(2x+3\right)\left(2x+b+3\right)}{-1}$$

Divide -\left(3+2x\right)\left(3+2x+b\right) by -1.

$$c=\left(2x+3\right)\left(2x+b+3\right)$$