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

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

Solve for m

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

$$x\neq \frac{1}{4}$$

Steps for Solving Linear Equation

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

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

Subtract 2 from both sides.

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

Combine all terms containing m.

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

The equation is in standard form.

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

Divide both sides by -4x+1.

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

Dividing by -4x+1 undoes the multiplication by -4x+1.

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

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

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

Solve for x (complex solution)

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

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

Solve for x

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

$$x=\frac{-\sqrt{\left(m-2\right)\left(m+1\right)}+m}{2}\text{, }m\leq -1\text{ or }m\geq 2$$