Solve (y-3)^2=10(x-4)

$$( y - 3 ) ^ { 2 } = 10 ( x - 4 )$$

Solve for x

$$x=\frac{\left(y-3\right)^{2}+40}{10}$$

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}.

$$y^{2}-6y+9=10\left(x-4\right)$$

Use the distributive property to multiply 10 by x-4.

$$y^{2}-6y+9=10x-40$$

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

$$10x-40=y^{2}-6y+9$$

Add 40 to both sides.

$$10x=y^{2}-6y+9+40$$

Add 9 and 40 to get 49.

$$10x=y^{2}-6y+49$$

Divide both sides by 10.

$$\frac{10x}{10}=\frac{y^{2}-6y+49}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x=\frac{y^{2}-6y+49}{10}$$

Divide y^{2}-6y+49 by 10.

$$x=\frac{y^{2}}{10}-\frac{3y}{5}+\frac{49}{10}$$

Solve for y (complex solution)

$$y=-\sqrt{10\left(x-4\right)}+3$$

$$y=\sqrt{10\left(x-4\right)}+3$$

Solve for y

$$y=-\sqrt{10\left(x-4\right)}+3$$

$$y=\sqrt{10\left(x-4\right)}+3\text{, }x\geq 4$$