Solve 4y^2+18x=21

$$4 y ^ { 2 } + 18 x = 21$$

Solve for x

$$x=-\frac{2y^{2}}{9}+\frac{7}{6}$$

Steps for Solving Linear Equation

Subtract 4y^{2} from both sides.

$$18x=21-4y^{2}$$

Divide both sides by 18.

$$\frac{18x}{18}=\frac{21-4y^{2}}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x=\frac{21-4y^{2}}{18}$$

Divide 21-4y^{2} by 18.

$$x=-\frac{2y^{2}}{9}+\frac{7}{6}$$

Solve for y (complex solution)

$$y=-\frac{\sqrt{21-18x}}{2}$$

$$y=\frac{\sqrt{21-18x}}{2}$$

Solve for y

$$y=\frac{\sqrt{21-18x}}{2}$$

$$y=-\frac{\sqrt{21-18x}}{2}\text{, }x\leq \frac{7}{6}$$