Solve sqrt{3}+6x^2+4y=0

$$\sqrt { 3 } + 6 x ^ { 2 } + 4 y = 0$$

Solve for y

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

Steps for Solving Linear Equation

Subtract \sqrt{3} from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}+4y=-\sqrt{3}$$

Subtract 6x^{2} from both sides.

$$4y=-\sqrt{3}-6x^{2}$$

The equation is in standard form.

$$4y=-6x^{2}-\sqrt{3}$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{-6x^{2}-\sqrt{3}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{-6x^{2}-\sqrt{3}}{4}$$

Divide -\sqrt{3}-6x^{2} by 4.

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{-24y-6\sqrt{3}}}{6}$$

$$x=\frac{\sqrt{-24y-6\sqrt{3}}}{6}$$

Solve for x

$$x=\frac{\sqrt{-24y-6\sqrt{3}}}{6}$$

$$x=-\frac{\sqrt{-24y-6\sqrt{3}}}{6}\text{, }y\leq -\frac{\sqrt{3}}{4}$$