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

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

Solve for k

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

$$x\neq 1$$

Steps for Solving Linear Equation

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

$$x^{2}-\left(kx+6x\right)+k+9=0$$

To find the opposite of kx+6x, find the opposite of each term.

$$x^{2}-kx-6x+k+9=0$$

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

$$-kx-6x+k+9=-x^{2}$$

Add 6x to both sides.

$$-kx+k+9=-x^{2}+6x$$

Subtract 9 from both sides.

$$-kx+k=-x^{2}+6x-9$$

Combine all terms containing k.

$$\left(-x+1\right)k=-x^{2}+6x-9$$

The equation is in standard form.

$$\left(1-x\right)k=-x^{2}+6x-9$$

Divide both sides by -x+1.

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

Dividing by -x+1 undoes the multiplication by -x+1.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{k\left(k+8\right)}+k+6}{2}$$

$$x=\frac{-\sqrt{k\left(k+8\right)}+k+6}{2}$$

Solve for x

$$x=\frac{\sqrt{k\left(k+8\right)}+k+6}{2}$$

$$x=\frac{-\sqrt{k\left(k+8\right)}+k+6}{2}\text{, }k\leq -8\text{ or }k\geq 0$$