Solve (x)=x+4+2xquadg:x-x^2

$$( x ) = x + 4 + 2 x \quad g : x - x ^ { 2 }$$

Solve for g

$$g=\frac{x^{2}-4}{2}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$xx=xx+x\times 4+2xg-x^{2}x$$

Multiply x and x to get x^{2}.

$$x^{2}=xx+x\times 4+2xg-x^{2}x$$

Multiply x and x to get x^{2}.

$$x^{2}=x^{2}+x\times 4+2xg-x^{2}x$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

$$x\times 4+2xg-x^{3}=0$$

Subtract x\times 4 from both sides. Anything subtracted from zero gives its negation.

$$2xg-x^{3}=-x\times 4$$

Add x^{3} to both sides.

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

Multiply -1 and 4 to get -4.

$$2xg=-4x+x^{3}$$

The equation is in standard form.

$$2xg=x^{3}-4x$$

Divide both sides by 2x.

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

Dividing by 2x undoes the multiplication by 2x.

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

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

$$g=\frac{x^{2}}{2}-2$$

Solve for x (complex solution)

$$x=-\sqrt{2\left(g+2\right)}$$

$$x=\sqrt{2\left(g+2\right)}\text{, }g\neq -2$$

Solve for x

$$x=\sqrt{2\left(g+2\right)}$$

$$x=-\sqrt{2\left(g+2\right)}\text{, }g>-2$$