Solve p-c=(x-2)^2

$$p - c = ( x - 2 ) ^ { 2 }$$

Solve for c

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

Steps for Solving Linear Equation

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

$$p-c=x^{2}-4x+4$$

Subtract p from both sides.

$$-c=x^{2}-4x+4-p$$

The equation is in standard form.

$$-c=x^{2}-4x-p+4$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

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

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

Solve for p

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

Solution Steps

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

$$p-c=x^{2}-4x+4$$

Add c to both sides.

$$p=x^{2}-4x+4+c$$