Solve x^2+8x-3x-10y+18=0

$$x ^ { 2 } + 8 x - 3 x - 10 y + 18 = 0$$

Solve for y

$$y=\frac{x^{2}+5x+18}{10}$$

Steps for Solving Linear Equation

Combine 8x and -3x to get 5x.

$$x^{2}+5x-10y+18=0$$

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

$$5x-10y+18=-x^{2}$$

Subtract 5x from both sides.

$$-10y+18=-x^{2}-5x$$

Subtract 18 from both sides.

$$-10y=-x^{2}-5x-18$$

Divide both sides by -10.

$$\frac{-10y}{-10}=\frac{-x^{2}-5x-18}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$y=\frac{-x^{2}-5x-18}{-10}$$

Divide -x^{2}-5x-18 by -10.

$$y=\frac{x^{2}}{10}+\frac{x}{2}+\frac{9}{5}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{40y-47}-5}{2}$$

$$x=\frac{-\sqrt{40y-47}-5}{2}$$

Solve for x

$$x=\frac{\sqrt{40y-47}-5}{2}$$

$$x=\frac{-\sqrt{40y-47}-5}{2}\text{, }y\geq \frac{47}{40}$$