Solve x(x+q-x/2)+8=0

$$x ( x + \frac { q - x } { 2 } ) + 8 = 0$$

Solve for q

$$q=-x-\frac{16}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2x\left(x+\frac{q-x}{2}\right)+16=0$$

Use the distributive property to multiply 2x by x+\frac{q-x}{2}.

$$2x^{2}+2x\times \frac{q-x}{2}+16=0$$

Express 2\times \frac{q-x}{2} as a single fraction.

$$2x^{2}+\frac{2\left(q-x\right)}{2}x+16=0$$

Cancel out 2 and 2.

$$2x^{2}+\left(q-x\right)x+16=0$$

Use the distributive property to multiply q-x by x.

$$2x^{2}+qx-x^{2}+16=0$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+qx+16=0$$

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

$$qx+16=-x^{2}$$

Subtract 16 from both sides.

$$qx=-x^{2}-16$$

The equation is in standard form.

$$xq=-x^{2}-16$$

Divide both sides by x.

$$\frac{xq}{x}=\frac{-x^{2}-16}{x}$$

Dividing by x undoes the multiplication by x.

$$q=\frac{-x^{2}-16}{x}$$

Divide -x^{2}-16 by x.

$$q=-x-\frac{16}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{q^{2}-64}-q}{2}$$

$$x=\frac{-\sqrt{q^{2}-64}-q}{2}$$

Solve for x

$$x=\frac{\sqrt{q^{2}-64}-q}{2}$$

$$x=\frac{-\sqrt{q^{2}-64}-q}{2}\text{, }|q|\geq 8$$