Solve +2x^2+x(m-16)+32-2m=0

$$+ 2 x ^ { 2 } + x ( m - 16 ) + 32 - 2 m = 0$$

Solve for m

$$m=-\frac{2\left(x-4\right)^{2}}{x-2}$$

$$x\neq 2$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by m-16.

$$2x^{2}+xm-16x+32-2m=0$$

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

$$xm-16x+32-2m=-2x^{2}$$

Add 16x to both sides.

$$xm+32-2m=-2x^{2}+16x$$

Subtract 32 from both sides.

$$xm-2m=-2x^{2}+16x-32$$

Combine all terms containing m.

$$\left(x-2\right)m=-2x^{2}+16x-32$$

Divide both sides by x-2.

$$\frac{\left(x-2\right)m}{x-2}=-\frac{2\left(x-4\right)^{2}}{x-2}$$

Dividing by x-2 undoes the multiplication by x-2.

$$m=-\frac{2\left(x-4\right)^{2}}{x-2}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{m\left(m-16\right)}-m+16}{4}$$

$$x=\frac{-\sqrt{m\left(m-16\right)}-m+16}{4}$$

Solve for x

$$x=\frac{\sqrt{m\left(m-16\right)}-m+16}{4}$$

$$x=\frac{-\sqrt{m\left(m-16\right)}-m+16}{4}\text{, }m\geq 16\text{ or }m\leq 0$$