Solve -9k^2+(6n+6)k-3n=0

$$- 9 k ^ { 2 } + ( 6 n + 6 ) k - 3 n = 0$$

Solve for n

$$n=-\frac{k\left(2-3k\right)}{2k-1}$$

$$k\neq \frac{1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 6n+6 by k.

$$-9k^{2}+6nk+6k-3n=0$$

Add 9k^{2} to both sides. Anything plus zero gives itself.

$$6nk+6k-3n=9k^{2}$$

Subtract 6k from both sides.

$$6nk-3n=9k^{2}-6k$$

Combine all terms containing n.

$$\left(6k-3\right)n=9k^{2}-6k$$

Divide both sides by 6k-3.

$$\frac{\left(6k-3\right)n}{6k-3}=\frac{3k\left(3k-2\right)}{6k-3}$$

Dividing by 6k-3 undoes the multiplication by 6k-3.

$$n=\frac{3k\left(3k-2\right)}{6k-3}$$

Divide 3k\left(-2+3k\right) by 6k-3.

$$n=\frac{k\left(3k-2\right)}{2k-1}$$

Solve for k

$$k=\frac{-\sqrt{n^{2}-n+1}+n+1}{3}$$

$$k=\frac{\sqrt{n^{2}-n+1}+n+1}{3}$$