Solve x^2-2nx+3q^2-b^2-b-2q+b=0

$$x ^ { 2 } - 2 n x + 3 q ^ { 2 } - b ^ { 2 } - b - 2 q + b = 0$$

Solve for n (complex solution)

$$\left\{\begin{matrix}n=\frac{x^{2}+3q^{2}-b^{2}-2q}{2x}\text{, }&x\neq 0\\n\in \mathrm{C}\text{, }&\left(q=\frac{\sqrt{3b^{2}+1}+1}{3}\text{ or }q=\frac{-\sqrt{3b^{2}+1}+1}{3}\right)\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine -b and b to get 0.

$$x^{2}-2nx+3q^{2}-b^{2}-2q=0$$

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

$$-2nx+3q^{2}-b^{2}-2q=-x^{2}$$

Subtract 3q^{2} from both sides.

$$-2nx-b^{2}-2q=-x^{2}-3q^{2}$$

Add b^{2} to both sides.

$$-2nx-2q=-x^{2}-3q^{2}+b^{2}$$

Add 2q to both sides.

$$-2nx=-x^{2}-3q^{2}+b^{2}+2q$$

The equation is in standard form.

$$\left(-2x\right)n=2q-3q^{2}+b^{2}-x^{2}$$

Divide both sides by -2x.

$$\frac{\left(-2x\right)n}{-2x}=\frac{2q-3q^{2}+b^{2}-x^{2}}{-2x}$$

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

$$n=\frac{2q-3q^{2}+b^{2}-x^{2}}{-2x}$$

Divide -x^{2}-3q^{2}+b^{2}+2q by -2x.

$$n=-\frac{2q-3q^{2}+b^{2}-x^{2}}{2x}$$

Solve for n

$$\left\{\begin{matrix}n=\frac{x^{2}+3q^{2}-b^{2}-2q}{2x}\text{, }&x\neq 0\\n\in \mathrm{R}\text{, }&\left(q=\frac{\sqrt{3b^{2}+1}+1}{3}\text{ or }q=\frac{-\sqrt{3b^{2}+1}+1}{3}\right)\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine -b and b to get 0.

$$x^{2}-2nx+3q^{2}-b^{2}-2q=0$$

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

$$-2nx+3q^{2}-b^{2}-2q=-x^{2}$$

Subtract 3q^{2} from both sides.

$$-2nx-b^{2}-2q=-x^{2}-3q^{2}$$

Add b^{2} to both sides.

$$-2nx-2q=-x^{2}-3q^{2}+b^{2}$$

Add 2q to both sides.

$$-2nx=-x^{2}-3q^{2}+b^{2}+2q$$

The equation is in standard form.

$$\left(-2x\right)n=2q-3q^{2}+b^{2}-x^{2}$$

Divide both sides by -2x.

$$\frac{\left(-2x\right)n}{-2x}=\frac{2q-3q^{2}+b^{2}-x^{2}}{-2x}$$

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

$$n=\frac{2q-3q^{2}+b^{2}-x^{2}}{-2x}$$

Divide -x^{2}-3q^{2}+b^{2}+2q by -2x.

$$n=-\frac{2q-3q^{2}+b^{2}-x^{2}}{2x}$$

Solve for b (complex solution)

$$b=-\sqrt{x^{2}-2nx+3q^{2}-2q}$$

$$b=\sqrt{x^{2}-2nx+3q^{2}-2q}$$

Solve for b

$$b=\sqrt{x^{2}-2nx+3q^{2}-2q}$$

$$b=-\sqrt{x^{2}-2nx+3q^{2}-2q}\text{, }x\leq -\sqrt{n^{2}-3q^{2}+2q}+n\text{ or }x\geq \sqrt{n^{2}-3q^{2}+2q}+n\text{ or }\left(n>-\sqrt{-\left(2q-3q^{2}\right)}\text{ and }q\geq \frac{2}{3}\text{ and }|n|<\sqrt{-\left(2q-3q^{2}\right)}\right)\text{ or }\left(n>-\sqrt{-\left(2q-3q^{2}\right)}\text{ and }q\leq 0\text{ and }|n|<\sqrt{-\left(2q-3q^{2}\right)}\right)\text{ or }\left(q\leq 0\text{ and }|n|\leq \sqrt{-\left(2q-3q^{2}\right)}\right)\text{ or }\left(q\geq \frac{2}{3}\text{ and }|n|\leq \sqrt{-\left(2q-3q^{2}\right)}\right)$$