Solve d^2+dx-x=0

$$d ^ { 2 } + d x - x = 0$$

Solve for x

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

$$d\neq 1$$

Steps for Solving Linear Equation

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

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

Combine all terms containing x.

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

Divide both sides by d-1.

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

Dividing by d-1 undoes the multiplication by d-1.

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

Solve for d (complex solution)

$$d=\frac{\sqrt{x\left(x+4\right)}-x}{2}$$

$$d=\frac{-\sqrt{x\left(x+4\right)}-x}{2}$$

Solve for d

$$d=\frac{\sqrt{x\left(x+4\right)}-x}{2}$$

$$d=\frac{-\sqrt{x\left(x+4\right)}-x}{2}\text{, }x\leq -4\text{ or }x\geq 0$$