Solve x^2-nx+n+4=0

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

Solve for n

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

$$x\neq 1$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-nx+n+4=-x^{2}$$

Subtract 4 from both sides.

$$-nx+n=-x^{2}-4$$

Combine all terms containing n.

$$\left(-x+1\right)n=-x^{2}-4$$

The equation is in standard form.

$$\left(1-x\right)n=-x^{2}-4$$

Divide both sides by -x+1.

$$\frac{\left(1-x\right)n}{1-x}=\frac{-x^{2}-4}{1-x}$$

Dividing by -x+1 undoes the multiplication by -x+1.

$$n=\frac{-x^{2}-4}{1-x}$$

Divide -x^{2}-4 by -x+1.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{n^{2}-4n-16}+n}{2}$$

$$x=\frac{-\sqrt{n^{2}-4n-16}+n}{2}$$

Solve for x

$$x=\frac{\sqrt{n^{2}-4n-16}+n}{2}$$

$$x=\frac{-\sqrt{n^{2}-4n-16}+n}{2}\text{, }n\geq 2\sqrt{5}+2\text{ or }n\leq 2-2\sqrt{5}$$