Solve 5x^2+2sqrt{y}=3

$$5 x ^ { 2 } + 2 \sqrt { y } = 3$$

Solve for y

$$y=\frac{\left(3-5x^{2}\right)^{2}}{4}$$

$$\frac{3-5x^{2}}{2}\geq 0$$

Steps for Solving Radical Equations

Subtract 5x^{2} from both sides of the equation.

$$2\sqrt{y}+5x^{2}-5x^{2}=3-5x^{2}$$

Subtracting 5x^{2} from itself leaves 0.

$$2\sqrt{y}=3-5x^{2}$$

Divide both sides by 2.

$$\frac{2\sqrt{y}}{2}=\frac{3-5x^{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$\sqrt{y}=\frac{3-5x^{2}}{2}$$

Square both sides of the equation.

$$y=\frac{\left(3-5x^{2}\right)^{2}}{4}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{-10\sqrt{y}+15}}{5}$$

$$x=\frac{\sqrt{-10\sqrt{y}+15}}{5}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{\left(3-5x^{2}\right)^{2}}{4}\text{, }&arg(\frac{3-5x^{2}}{2})<\pi \\y=0\text{, }&x=-\frac{\sqrt{15}}{5}\text{ or }x=\frac{\sqrt{15}}{5}\end{matrix}\right.$$

Solve for x

$$x=\frac{\sqrt{-10\sqrt{y}+15}}{5}$$

$$x=-\frac{\sqrt{-10\sqrt{y}+15}}{5}\text{, }y\geq 0\text{ and }y\leq \frac{9}{4}$$