Solve 3x^2+2mx+m+6=0

$$3 x ^ { 2 } + 2 m x + m + 6 = 0$$

Solve for m

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

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

Steps for Solving Linear Equation

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

$$2mx+m+6=-3x^{2}$$

Subtract 6 from both sides.

$$2mx+m=-3x^{2}-6$$

Combine all terms containing m.

$$\left(2x+1\right)m=-3x^{2}-6$$

Divide both sides by 2x+1.

$$\frac{\left(2x+1\right)m}{2x+1}=\frac{-3x^{2}-6}{2x+1}$$

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

$$m=\frac{-3x^{2}-6}{2x+1}$$

Divide -3x^{2}-6 by 2x+1.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(m-6\right)\left(m+3\right)}-m}{3}$$

$$x=\frac{-\sqrt{\left(m-6\right)\left(m+3\right)}-m}{3}$$

Solve for x

$$x=\frac{\sqrt{\left(m-6\right)\left(m+3\right)}-m}{3}$$

$$x=\frac{-\sqrt{\left(m-6\right)\left(m+3\right)}-m}{3}\text{, }m\leq -3\text{ or }m\geq 6$$