Solve 3x^2+(k+2)x+k+2=0

$$3 x ^ { 2 } + ( k + 2 ) x + k + 2 = 0$$

Solve for k

$$k=-\frac{3x^{2}+2x+2}{x+1}$$

$$x\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply k+2 by x.

$$3x^{2}+kx+2x+k+2=0$$

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$kx+2x+k+2=-3x^{2}$$

Subtract 2x from both sides.

$$kx+k+2=-3x^{2}-2x$$

Subtract 2 from both sides.

$$kx+k=-3x^{2}-2x-2$$

Combine all terms containing k.

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

Divide both sides by x+1.

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

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

$$k=\frac{-3x^{2}-2x-2}{x+1}$$

Divide -3x^{2}-2x-2 by x+1.

$$k=-\frac{3x^{2}+2x+2}{x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-10\right)\left(k+2\right)}-k-2}{6}$$

$$x=\frac{-\sqrt{\left(k-10\right)\left(k+2\right)}-k-2}{6}$$

Solve for x

$$x=\frac{\sqrt{\left(k-10\right)\left(k+2\right)}-k-2}{6}$$

$$x=\frac{-\sqrt{\left(k-10\right)\left(k+2\right)}-k-2}{6}\text{, }k\leq -2\text{ or }k\geq 10$$