Solve n4x^2-32x+64=0

$$n 4 x ^ { 2 } - 32 x + 64 = 0$$

Solve for n_4

$$n_{4}=-\frac{32\left(2-x\right)}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Add 32x to both sides. Anything plus zero gives itself.

$$n_{4}x^{2}+64=32x$$

Subtract 64 from both sides.

$$n_{4}x^{2}=32x-64$$

The equation is in standard form.

$$x^{2}n_{4}=32x-64$$

Divide both sides by x^{2}.

$$\frac{x^{2}n_{4}}{x^{2}}=\frac{32x-64}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$n_{4}=\frac{32x-64}{x^{2}}$$

Divide -64+32x by x^{2}.

$$n_{4}=\frac{32\left(x-2\right)}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{8\left(\sqrt{4-n_{4}}+2\right)}{n_{4}}\text{; }x=\frac{8\left(-\sqrt{4-n_{4}}+2\right)}{n_{4}}\text{, }&n_{4}\neq 0\\x=2\text{, }&n_{4}=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{8\left(\sqrt{4-n_{4}}+2\right)}{n_{4}}\text{; }x=\frac{8\left(-\sqrt{4-n_{4}}+2\right)}{n_{4}}\text{, }&n_{4}\neq 0\text{ and }n_{4}\leq 4\\x=2\text{, }&n_{4}=0\end{matrix}\right.$$