Solve 2x^2-2(p-4)x+p=0

$$2 x ^ { 2 } - 2 ( p - 4 ) x + p = 0$$

Solve for p

$$p=-\frac{2x\left(x+4\right)}{1-2x}$$

$$x\neq \frac{1}{2}$$

Steps for Solving Linear Equation

Multiply -1 and 2 to get -2.

$$2x^{2}-2\left(p-4\right)x+p=0$$

Use the distributive property to multiply -2 by p-4.

$$2x^{2}+\left(-2p+8\right)x+p=0$$

Use the distributive property to multiply -2p+8 by x.

$$2x^{2}-2px+8x+p=0$$

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

$$-2px+8x+p=-2x^{2}$$

Subtract 8x from both sides.

$$-2px+p=-2x^{2}-8x$$

Combine all terms containing p.

$$\left(-2x+1\right)p=-2x^{2}-8x$$

The equation is in standard form.

$$\left(1-2x\right)p=-2x^{2}-8x$$

Divide both sides by -2x+1.

$$\frac{\left(1-2x\right)p}{1-2x}=-\frac{2x\left(x+4\right)}{1-2x}$$

Dividing by -2x+1 undoes the multiplication by -2x+1.

$$p=-\frac{2x\left(x+4\right)}{1-2x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(p-8\right)\left(p-2\right)}+p-4}{2}$$

$$x=\frac{-\sqrt{\left(p-8\right)\left(p-2\right)}+p-4}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(p-8\right)\left(p-2\right)}+p-4}{2}$$

$$x=\frac{-\sqrt{\left(p-8\right)\left(p-2\right)}+p-4}{2}\text{, }p\leq 2\text{ or }p\geq 8$$