Solve left(2x+3right)^2+y(2x+3)-z=0

$${ \left(2x+3 \right) }^{ 2 } +y(2x+3)-z=0$$

Solve for y (complex solution)

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

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}.

$$4x^{2}+12x+9+y\left(2x+3\right)-z=0$$

Use the distributive property to multiply y by 2x+3.

$$4x^{2}+12x+9+2yx+3y-z=0$$

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

$$12x+9+2yx+3y-z=-4x^{2}$$

Subtract 12x from both sides.

$$9+2yx+3y-z=-4x^{2}-12x$$

Subtract 9 from both sides.

$$2yx+3y-z=-4x^{2}-12x-9$$

Add z to both sides.

$$2yx+3y=-4x^{2}-12x-9+z$$

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by 2x+3.

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

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

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

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

$$y=\frac{-4x^{2}-12x+z-9}{2x+3}$$

Solve for y

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

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}.

$$4x^{2}+12x+9+y\left(2x+3\right)-z=0$$

Use the distributive property to multiply y by 2x+3.

$$4x^{2}+12x+9+2yx+3y-z=0$$

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

$$12x+9+2yx+3y-z=-4x^{2}$$

Subtract 12x from both sides.

$$9+2yx+3y-z=-4x^{2}-12x$$

Subtract 9 from both sides.

$$2yx+3y-z=-4x^{2}-12x-9$$

Add z to both sides.

$$2yx+3y=-4x^{2}-12x-9+z$$

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by 2x+3.

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

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

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

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

$$y=\frac{-4x^{2}-12x+z-9}{2x+3}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}+4z}}{4}-\frac{y}{4}-\frac{3}{2}$$

$$x=-\frac{\sqrt{y^{2}+4z}}{4}-\frac{y}{4}-\frac{3}{2}$$

Solve for x

$$x=-\frac{\sqrt{y^{2}+4z}}{4}-\frac{y}{4}-\frac{3}{2}$$

$$x=\frac{\sqrt{y^{2}+4z}}{4}-\frac{y}{4}-\frac{3}{2}\text{, }z\geq -\frac{y^{2}}{4}$$