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

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

Solve for p

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

Steps for Solving Linear Equation

Use the distributive property to multiply 2x^{2}-4 by x.

$$-x^{2}+2x^{3}-4x-p=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$2x^{3}-4x-p=x^{2}$$

Subtract 2x^{3} from both sides.

$$-4x-p=x^{2}-2x^{3}$$

Add 4x to both sides.

$$-p=x^{2}-2x^{3}+4x$$

The equation is in standard form.

$$-p=4x+x^{2}-2x^{3}$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide x\left(x-2x^{2}+4\right) by -1.

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