Solve 4N=(9-x)^2+(2sqrt{2})^2

$$4 N = ( 9 - x ) ^ { 2 } + ( 2 \sqrt { 2 } ) ^ { 2 }$$

Solve for N

$$N=\frac{\left(9-x\right)^{2}+8}{4}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$4N=81-18x+x^{2}+\left(2\sqrt{2}\right)^{2}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$4N=81-18x+x^{2}+2^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4N=81-18x+x^{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4N=81-18x+x^{2}+4\times 2$$

Multiply 4 and 2 to get 8.

$$4N=81-18x+x^{2}+8$$

Add 81 and 8 to get 89.

$$4N=89-18x+x^{2}$$

The equation is in standard form.

$$4N=x^{2}-18x+89$$

Divide both sides by 4.

$$\frac{4N}{4}=\frac{x^{2}-18x+89}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$N=\frac{x^{2}-18x+89}{4}$$

Divide 89-18x+x^{2} by 4.

$$N=\frac{x^{2}}{4}-\frac{9x}{2}+\frac{89}{4}$$

Solve for x (complex solution)

$$x=2\sqrt{N-2}+9$$

$$x=-2\sqrt{N-2}+9$$

Solve for x

$$x=2\sqrt{N-2}+9$$

$$x=-2\sqrt{N-2}+9\text{, }N\geq 2$$