Solve 7x^2+8x-3y+7=0

$$7 x ^ { 2 } + 8 x - 3 y + 7 = 0$$

Solve for y

$$y=\frac{7x^{2}+8x+7}{3}$$

Steps for Solving Linear Equation

Subtract 7x^{2} from both sides. Anything subtracted from zero gives its negation.

$$8x-3y+7=-7x^{2}$$

Subtract 8x from both sides.

$$-3y+7=-7x^{2}-8x$$

Subtract 7 from both sides.

$$-3y=-7x^{2}-8x-7$$

Divide both sides by -3.

$$\frac{-3y}{-3}=\frac{-7x^{2}-8x-7}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y=\frac{-7x^{2}-8x-7}{-3}$$

Divide -7x^{2}-8x-7 by -3.

$$y=\frac{7x^{2}+8x+7}{3}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{21y-33}-4}{7}$$

$$x=\frac{-\sqrt{21y-33}-4}{7}$$

Solve for x

$$x=\frac{\sqrt{21y-33}-4}{7}$$

$$x=\frac{-\sqrt{21y-33}-4}{7}\text{, }y\geq \frac{11}{7}$$