Solve 2x+sqrt{3y}=x^2-4

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

Solve for y

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

$$-\left(4+2x-x^{2}\right)\geq 0$$

Steps for Solving Radical Equations

Subtract 2x from both sides of the equation.

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

Subtracting 2x from itself leaves 0.

$$\sqrt{3y}=x^{2}-4-2x$$

Subtract 2x from x^{2}-4.

$$\sqrt{3y}=x^{2}-2x-4$$

Square both sides of the equation.

$$3y=\left(x^{2}-2x-4\right)^{2}$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Solve for x (complex solution)

$$x=-\sqrt{\sqrt{3y}+5}+1$$

$$x=\sqrt{\sqrt{3y}+5}+1$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{\left(4+2x-x^{2}\right)^{2}}{3}\text{, }&arg(4+2x-x^{2})\geq \pi \\y=0\text{, }&x=\sqrt{5}+1\text{ or }x=1-\sqrt{5}\end{matrix}\right.$$

Solve for x

$$x=-\sqrt{\sqrt{3y}+5}+1$$

$$x=\sqrt{\sqrt{3y}+5}+1\text{, }y\geq 0$$