Solve (x-5)^2-c=0

$$( x - 5 ) ^ { 2 } - c = 0$$

Solve for c

$$c=\left(x-5\right)^{2}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-c=0$$

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

$$-10x+25-c=-x^{2}$$

Add 10x to both sides.

$$25-c=-x^{2}+10x$$

Subtract 25 from both sides.

$$-c=-x^{2}+10x-25$$

Divide both sides by -1.

$$\frac{-c}{-1}=-\frac{\left(x-5\right)^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$c=-\frac{\left(x-5\right)^{2}}{-1}$$

Divide -\left(x-5\right)^{2} by -1.

$$c=\left(x-5\right)^{2}$$

Solve for x (complex solution)

$$x=-\sqrt{c}+5$$

$$x=\sqrt{c}+5$$

Solve for x

$$x=-\sqrt{c}+5$$

$$x=\sqrt{c}+5\text{, }c\geq 0$$