Solve 3x^2-(2m+12)x+216=0

$$3 x ^ { 2 } - ( 2 m + 12 ) x + 216 = 0$$

Solve for m

$$m=\frac{3x}{2}-6+\frac{108}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2m+12 by x.

$$3x^{2}-\left(2mx+12x\right)+216=0$$

To find the opposite of 2mx+12x, find the opposite of each term.

$$3x^{2}-2mx-12x+216=0$$

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

$$-2mx-12x+216=-3x^{2}$$

Add 12x to both sides.

$$-2mx+216=-3x^{2}+12x$$

Subtract 216 from both sides.

$$-2mx=-3x^{2}+12x-216$$

The equation is in standard form.

$$\left(-2x\right)m=-3x^{2}+12x-216$$

Divide both sides by -2x.

$$\frac{\left(-2x\right)m}{-2x}=\frac{-3x^{2}+12x-216}{-2x}$$

Dividing by -2x undoes the multiplication by -2x.

$$m=\frac{-3x^{2}+12x-216}{-2x}$$

Divide 12x-3x^{2}-216 by -2x.

$$m=\frac{3x}{2}-6+\frac{108}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{m^{2}+12m-612}+m+6}{3}$$

$$x=\frac{-\sqrt{m^{2}+12m-612}+m+6}{3}$$

Solve for x

$$x=\frac{\sqrt{m^{2}+12m-612}+m+6}{3}$$

$$x=\frac{-\sqrt{m^{2}+12m-612}+m+6}{3}\text{, }m\geq 18\sqrt{2}-6\text{ or }m\leq -18\sqrt{2}-6$$