Solve 2x^2+px+p+6=0

$$2 { x }^{ 2 } +px+p+6 = 0$$

Solve for p

$$p=-\frac{2\left(x^{2}+3\right)}{x+1}$$

$$x\neq -1$$

Steps for Solving Linear Equation

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

$$px+p+6=-2x^{2}$$

Subtract 6 from both sides.

$$px+p=-2x^{2}-6$$

Combine all terms containing p.

$$\left(x+1\right)p=-2x^{2}-6$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)p}{x+1}=\frac{-2x^{2}-6}{x+1}$$

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

$$p=\frac{-2x^{2}-6}{x+1}$$

Divide -2x^{2}-6 by x+1.

$$p=-\frac{2\left(x^{2}+3\right)}{x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(p-12\right)\left(p+4\right)}-p}{4}$$

$$x=\frac{-\sqrt{\left(p-12\right)\left(p+4\right)}-p}{4}$$

Solve for x

$$x=\frac{\sqrt{\left(p-12\right)\left(p+4\right)}-p}{4}$$

$$x=\frac{-\sqrt{\left(p-12\right)\left(p+4\right)}-p}{4}\text{, }p\leq -4\text{ or }p\geq 12$$