Solve x^2-(p-2)x+(2p+1)=0

$$x ^ { 2 } - ( p - 2 ) x + ( 2 p + 1 ) = 0$$

Solve for p

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

$$x\neq 2$$

Steps for Solving Linear Equation

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

$$x^{2}-\left(px-2x\right)+2p+1=0$$

To find the opposite of px-2x, find the opposite of each term.

$$x^{2}-px+2x+2p+1=0$$

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

$$-px+2x+2p+1=-x^{2}$$

Subtract 2x from both sides.

$$-px+2p+1=-x^{2}-2x$$

Subtract 1 from both sides.

$$-px+2p=-x^{2}-2x-1$$

Combine all terms containing p.

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

The equation is in standard form.

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

Divide both sides by -x+2.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{p\left(p-12\right)}+p-2}{2}$$

$$x=\frac{-\sqrt{p\left(p-12\right)}+p-2}{2}$$

Solve for x

$$x=\frac{\sqrt{p\left(p-12\right)}+p-2}{2}$$

$$x=\frac{-\sqrt{p\left(p-12\right)}+p-2}{2}\text{, }p\geq 12\text{ or }p\leq 0$$