Solve g(x)=(x-2)^2+5quadn^primey_{1}7

$$g ( x ) = ( x - 2 ) ^ { 2 } + 5 \quad n ^ { \prime } y _ { 1 } 7$$

Solve for g

$$g=\frac{\left(x-2\right)^{2}}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$gx=x^{2}-4x+4+5\frac{\mathrm{d}}{\mathrm{d}x}(n)y_{1}\times 7$$

Multiply 5 and 7 to get 35.

$$gx=x^{2}-4x+4+35\frac{\mathrm{d}}{\mathrm{d}x}(n)y_{1}$$

The equation is in standard form.

$$xg=x^{2}-4x+4$$

Divide both sides by x.

$$\frac{xg}{x}=\frac{\left(x-2\right)^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$g=\frac{\left(x-2\right)^{2}}{x}$$

Solve for n

$$n\in \mathrm{R}$$

$$g=\frac{\left(x-2\right)^{2}}{x}\text{ and }x\neq 0$$