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

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

Solve for y

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

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2} by y-3.

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

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

$$\frac{1}{2}y-\frac{3}{2}=x^{2}-4x+4$$

Add \frac{3}{2} to both sides.

$$\frac{1}{2}y=x^{2}-4x+4+\frac{3}{2}$$

Add 4 and \frac{3}{2} to get \frac{11}{2}.

$$\frac{1}{2}y=x^{2}-4x+\frac{11}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}y}{\frac{1}{2}}=\frac{x^{2}-4x+\frac{11}{2}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$y=\frac{x^{2}-4x+\frac{11}{2}}{\frac{1}{2}}$$

Divide x^{2}-4x+\frac{11}{2} by \frac{1}{2} by multiplying x^{2}-4x+\frac{11}{2} by the reciprocal of \frac{1}{2}.

$$y=2x^{2}-8x+11$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{2y-6}}{2}+2$$

$$x=\frac{\sqrt{2y-6}}{2}+2$$

Solve for x

$$x=-\frac{\sqrt{2y-6}}{2}+2$$

$$x=\frac{\sqrt{2y-6}}{2}+2\text{, }y\geq 3$$