Solve 4x^2+2mx+m+3=0

$$4 x ^ { 2 } + 2 m x + m + 3 = 0$$

Solve for m

$$m=-\frac{4x^{2}+3}{2x+1}$$

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

Steps for Solving Linear Equation

Subtract 4x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2mx+m+3=-4x^{2}$$

Subtract 3 from both sides.

$$2mx+m=-4x^{2}-3$$

Combine all terms containing m.

$$\left(2x+1\right)m=-4x^{2}-3$$

Divide both sides by 2x+1.

$$\frac{\left(2x+1\right)m}{2x+1}=\frac{-4x^{2}-3}{2x+1}$$

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

$$m=\frac{-4x^{2}-3}{2x+1}$$

Divide -4x^{2}-3 by 2x+1.

$$m=-\frac{4x^{2}+3}{2x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(m-6\right)\left(m+2\right)}-m}{4}$$

$$x=\frac{-\sqrt{\left(m-6\right)\left(m+2\right)}-m}{4}$$

Solve for x

$$x=\frac{\sqrt{\left(m-6\right)\left(m+2\right)}-m}{4}$$

$$x=\frac{-\sqrt{\left(m-6\right)\left(m+2\right)}-m}{4}\text{, }m\leq -2\text{ or }m\geq 6$$