Solve (x-1)^2=-y+10/3

$$( x - 1 ) ^ { 2 } = - y + \frac { 10 } { 3 }$$

Solve for y

$$y=-\left(x-1\right)^{2}+\frac{10}{3}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1=-y+\frac{10}{3}$$

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

$$-y+\frac{10}{3}=x^{2}-2x+1$$

Subtract \frac{10}{3} from both sides.

$$-y=x^{2}-2x+1-\frac{10}{3}$$

Subtract \frac{10}{3} from 1 to get -\frac{7}{3}.

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

Divide both sides by -1.

$$\frac{-y}{-1}=\frac{x^{2}-2x-\frac{7}{3}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=\frac{x^{2}-2x-\frac{7}{3}}{-1}$$

Divide x^{2}-2x-\frac{7}{3} by -1.

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{30-9y}}{3}+1$$

$$x=\frac{\sqrt{30-9y}}{3}+1$$

Solve for x

$$x=-\frac{\sqrt{30-9y}}{3}+1$$

$$x=\frac{\sqrt{30-9y}}{3}+1\text{, }y\leq \frac{10}{3}$$