Solve left(y+4right)^2=12(x-1)

$${ \left(y+4 \right) }^{ 2 } =12(x-1)$$

Solve for x

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.

$$y^{2}+8y+16=12\left(x-1\right)$$

Use the distributive property to multiply 12 by x-1.

$$y^{2}+8y+16=12x-12$$

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

$$12x-12=y^{2}+8y+16$$

Add 12 to both sides.

$$12x=y^{2}+8y+16+12$$

Add 16 and 12 to get 28.

$$12x=y^{2}+8y+28$$

Divide both sides by 12.

$$\frac{12x}{12}=\frac{y^{2}+8y+28}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x=\frac{y^{2}+8y+28}{12}$$

Divide y^{2}+8y+28 by 12.

$$x=\frac{y^{2}}{12}+\frac{2y}{3}+\frac{7}{3}$$

Solve for y (complex solution)

$$y=-2\sqrt{3x-3}-4$$

$$y=2\sqrt{3x-3}-4$$

Solve for y

$$y=-2\sqrt{3x-3}-4$$

$$y=2\sqrt{3x-3}-4\text{, }x\geq 1$$