Solve -x^2-alpha+4=2(x^2-x-x+1)

$$- x ^ { 2 } - \alpha + 4 = 2 ( x ^ { 2 } - x - x + 1 )$$

Solve for α

$$\alpha =2+4x-3x^{2}$$

Steps for Solving Linear Equation

Combine -x and -x to get -2x.

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

Use the distributive property to multiply 2 by x^{2}-2x+1.

$$-x^{2}-\alpha +4=2x^{2}-4x+2$$

Add x^{2} to both sides.

$$-\alpha +4=2x^{2}-4x+2+x^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$-\alpha +4=3x^{2}-4x+2$$

Subtract 4 from both sides.

$$-\alpha =3x^{2}-4x+2-4$$

Subtract 4 from 2 to get -2.

$$-\alpha =3x^{2}-4x-2$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

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

$$\alpha =2+4x-3x^{2}$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{10-3\alpha }+2}{3}$$

$$x=\frac{\sqrt{10-3\alpha }+2}{3}$$

Solve for x

$$x=\frac{-\sqrt{10-3\alpha }+2}{3}$$

$$x=\frac{\sqrt{10-3\alpha }+2}{3}\text{, }\alpha \leq \frac{10}{3}$$