Solve (p+7)^2=p^2+q^2

$$( p + 7 ) ^ { 2 } = p ^ { 2 } + q ^ { 2 }$$

Solve for p

$$p=\frac{q^{2}-49}{14}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(p+7\right)^{2}.

$$p^{2}+14p+49=p^{2}+q^{2}$$

Subtract p^{2} from both sides.

$$p^{2}+14p+49-p^{2}=q^{2}$$

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

$$14p+49=q^{2}$$

Subtract 49 from both sides.

$$14p=q^{2}-49$$

Divide both sides by 14.

$$\frac{14p}{14}=\frac{q^{2}-49}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$p=\frac{q^{2}-49}{14}$$

Divide q^{2}-49 by 14.

$$p=\frac{q^{2}}{14}-\frac{7}{2}$$

Solve for q (complex solution)

$$q=-\sqrt{14p+49}$$

$$q=\sqrt{14p+49}$$

Solve for q

$$q=\sqrt{14p+49}$$

$$q=-\sqrt{14p+49}\text{, }p\geq -\frac{7}{2}$$