Solve (x-k)^2-(k-2)^2=6

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

Solve for k

$$k=-\frac{x^{2}-10}{2\left(2-x\right)}$$

$$x\neq 2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-k\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-2\right)^{2}.

$$x^{2}-2xk+k^{2}-\left(k^{2}-4k+4\right)=6$$

To find the opposite of k^{2}-4k+4, find the opposite of each term.

$$x^{2}-2xk+k^{2}-k^{2}+4k-4=6$$

Combine k^{2} and -k^{2} to get 0.

$$x^{2}-2xk+4k-4=6$$

Subtract x^{2} from both sides.

$$-2xk+4k-4=6-x^{2}$$

Add 4 to both sides.

$$-2xk+4k=6-x^{2}+4$$

Add 6 and 4 to get 10.

$$-2xk+4k=10-x^{2}$$

Combine all terms containing k.

$$\left(-2x+4\right)k=10-x^{2}$$

The equation is in standard form.

$$\left(4-2x\right)k=10-x^{2}$$

Divide both sides by -2x+4.

$$\frac{\left(4-2x\right)k}{4-2x}=\frac{10-x^{2}}{4-2x}$$

Dividing by -2x+4 undoes the multiplication by -2x+4.

$$k=\frac{10-x^{2}}{4-2x}$$

Divide 10-x^{2} by -2x+4.

$$k=\frac{10-x^{2}}{2\left(2-x\right)}$$

Solve for x

$$x=\sqrt{k^{2}-4k+10}+k$$

$$x=-\sqrt{k^{2}-4k+10}+k$$