Solve 3x^2+18x-y+24=0

$$3 x ^ { 2 } + 18 x - y + 24 = 0$$

Solve for y

$$y=3\left(x+2\right)\left(x+4\right)$$

Steps for Solving Linear Equation

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$18x-y+24=-3x^{2}$$

Subtract 18x from both sides.

$$-y+24=-3x^{2}-18x$$

Subtract 24 from both sides.

$$-y=-3x^{2}-18x-24$$

Divide both sides by -1.

$$\frac{-y}{-1}=-\frac{3\left(x+2\right)\left(x+4\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=-\frac{3\left(x+2\right)\left(x+4\right)}{-1}$$

Divide -3\left(2+x\right)\left(4+x\right) by -1.

$$y=3\left(x+2\right)\left(x+4\right)$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3y+9}}{3}-3$$

$$x=-\frac{\sqrt{3y+9}}{3}-3$$

Solve for x

$$x=\frac{\sqrt{3y+9}}{3}-3$$

$$x=-\frac{\sqrt{3y+9}}{3}-3\text{, }y\geq -3$$