Solve 2x^4+3x^3-nx^2+3x-9=0

$$2 x ^ { 4 } + 3 x ^ { 3 } - n x ^ { 2 } + 3 x - 9 = 0$$

Solve for n

$$n=\frac{2x^{4}+3x^{3}+3x-9}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 3x from both sides. Anything subtracted from zero gives its negation.

$$2x^{4}+3x^{3}-nx^{2}-9=-3x$$

Add 9 to both sides.

$$2x^{4}+3x^{3}-nx^{2}=-3x+9$$

Subtract 2x^{4} from both sides.

$$3x^{3}-nx^{2}=-3x+9-2x^{4}$$

Subtract 3x^{3} from both sides.

$$-nx^{2}=-3x+9-2x^{4}-3x^{3}$$

The equation is in standard form.

$$\left(-x^{2}\right)n=9-3x-3x^{3}-2x^{4}$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)n}{-x^{2}}=\frac{9-3x-3x^{3}-2x^{4}}{-x^{2}}$$

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

$$n=\frac{9-3x-3x^{3}-2x^{4}}{-x^{2}}$$

Divide -3x+9-2x^{4}-3x^{3} by -x^{2}.

$$n=2x^{2}+3x+\frac{3x-9}{x^{2}}$$