Solve x=2+dx^2/2!

$$x = 2 + \frac { d x ^ { 2 } } { 2 ! }$$

Solve for d

$$d=-\frac{2\left(2-x\right)}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

The factorial of 2 is 2.

$$x=2+\frac{dx^{2}}{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2+\frac{dx^{2}}{2}=x$$

Subtract 2 from both sides.

$$\frac{dx^{2}}{2}=x-2$$

Multiply both sides of the equation by 2.

$$dx^{2}=2x-4$$

The equation is in standard form.

$$x^{2}d=2x-4$$

Divide both sides by x^{2}.

$$\frac{x^{2}d}{x^{2}}=\frac{2x-4}{x^{2}}$$

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

$$d=\frac{2x-4}{x^{2}}$$

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

$$d=\frac{2\left(x-2\right)}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{1-4d}-1}{d}\text{; }x=\frac{\sqrt{1-4d}+1}{d}\text{, }&d\neq 0\\x=2\text{, }&d=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{1-4d}-1}{d}\text{; }x=\frac{\sqrt{1-4d}+1}{d}\text{, }&d\neq 0\text{ and }d\leq \frac{1}{4}\\x=2\text{, }&d=0\end{matrix}\right.$$