Solve for q
q=-\frac{2x\left(3x+1\right)}{3x-1}
x\neq \frac{1}{3}
Solve for x (complex solution)
x=\frac{\sqrt{9q^{2}+36q+4}}{12}-\frac{q}{4}-\frac{1}{6}
x=-\frac{\sqrt{9q^{2}+36q+4}}{12}-\frac{q}{4}-\frac{1}{6}
Solve for x
x=\frac{\sqrt{9q^{2}+36q+4}}{12}-\frac{q}{4}-\frac{1}{6}
x=-\frac{\sqrt{9q^{2}+36q+4}}{12}-\frac{q}{4}-\frac{1}{6}\text{, }q\geq \frac{4\sqrt{2}}{3}-2\text{ or }q\leq -\frac{4\sqrt{2}}{3}-2
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2x+3qx-q=-6x^{2}
Subtract 6x^{2} from both sides. Anything subtracted from zero gives its negation.
3qx-q=-6x^{2}-2x
Subtract 2x from both sides.
\left(3x-1\right)q=-6x^{2}-2x
Combine all terms containing q.
\frac{\left(3x-1\right)q}{3x-1}=-\frac{2x\left(3x+1\right)}{3x-1}
Divide both sides by 3x-1.
q=-\frac{2x\left(3x+1\right)}{3x-1}
Dividing by 3x-1 undoes the multiplication by 3x-1.
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