Solve (y+3)^2=20(x-1)

$$( y + 3 ) ^ { 2 } = 20 ( x - 1 )$$

Solve for x

$$x=\frac{\left(y+3\right)^{2}+20}{20}$$

Steps for Solving Linear Equation

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

$$y^{2}+6y+9=20\left(x-1\right)$$

Use the distributive property to multiply 20 by x-1.

$$y^{2}+6y+9=20x-20$$

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

$$20x-20=y^{2}+6y+9$$

Add 20 to both sides.

$$20x=y^{2}+6y+9+20$$

Add 9 and 20 to get 29.

$$20x=y^{2}+6y+29$$

Divide both sides by 20.

$$\frac{20x}{20}=\frac{y^{2}+6y+29}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x=\frac{y^{2}+6y+29}{20}$$

Divide y^{2}+6y+29 by 20.

$$x=\frac{y^{2}}{20}+\frac{3y}{10}+\frac{29}{20}$$

Solve for y (complex solution)

$$y=-2\sqrt{5\left(x-1\right)}-3$$

$$y=2\sqrt{5\left(x-1\right)}-3$$

Solve for y

$$y=-2\sqrt{5\left(x-1\right)}-3$$

$$y=2\sqrt{5\left(x-1\right)}-3\text{, }x\geq 1$$