Solve Dx=-1/2x^2

$$D x = - \frac { 1 } { 2 x ^ { 2 } }$$

Solve for D

$$D=-\frac{1}{2x^{3}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x^{2}.

$$Dx\times 2x^{2}=-1$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$Dx^{3}\times 2=-1$$

The equation is in standard form.

$$2x^{3}D=-1$$

Divide both sides by 2x^{3}.

$$\frac{2x^{3}D}{2x^{3}}=-\frac{1}{2x^{3}}$$

Dividing by 2x^{3} undoes the multiplication by 2x^{3}.

$$D=-\frac{1}{2x^{3}}$$

Solve for x (complex solution)

$$x=-\frac{2^{\frac{2}{3}}D^{-\frac{1}{3}}}{2}$$

$$x=\frac{2^{\frac{2}{3}}e^{\frac{\pi i}{3}}D^{-\frac{1}{3}}}{2}$$

$$x=\frac{2^{\frac{2}{3}}e^{\frac{5\pi i}{3}}D^{-\frac{1}{3}}}{2}\text{, }D\neq 0$$

Solve for x

$$x=-\frac{2^{\frac{2}{3}}}{2\sqrt[3]{D}}$$

$$D\neq 0$$