Completing the square:

Since the equation has x and y squared terms on the left and 0 on the right, the equation represents pair of lines crossing each other at \(\displaystyle{\left({4},-{2}\right)}.\)

\(\displaystyle{4}{x}^{2}-{y}^{2}-{32}{x}-{4}{y}+{60}={0}\)

\(\displaystyle{4}{x}^{2}-{32}{x}-{y}^{2}-{4}{y}=-{60}\)

\(\displaystyle{4}{\left({x}^{2}-{8}{x}\right)}-{\left({y}^{2}+{4}{y}\right)}=-{60}\)

\(\displaystyle{4}{\left({\left({x}\right)}^{2}-{2}{\left({x}\right)}{\left({4}\right)}+{4}^{2}-{4}^{2}\right)}-{\left({y}^{2}+{2}{\left({y}\right)}{\left({2}\right)}+{2}^{2}-{2}^{2}\right)}=-{60}\)

\(\displaystyle{4}{\left({\left({x}\right)}^{2}-{2}{\left({x}\right)}{\left({4}\right)}+{4}^{2}-{16}\right)}-{\left({y}^{2}\right)}+{2}{\left({y}\right)}{\left({2}\right)}+{2}^{2}-{4}=-{60}\)

\(\displaystyle{4}{\left({\left({x}-{4}\right)}^{2}-{16}\right)}-{\left({\left({y}+{2}\right)}^{2}-{4}\right)}=-{60}\)

\(\displaystyle{4}{\left({\left({x}-{4}\right)}^{2}\right)}-{64}-{\left({\left({y}+{2}\right)}^{2}\right)}+{4}=-{60}\)

\(\displaystyle{4}{\left({x}-{4}\right)}^{2}-{\left({y}+{2}\right)}^{2}={0}\)

Step 2

The equations of the lines are \(\displaystyle{2}{x}–{y}={10}{\quad\text{and}\quad}{2}{x}+{y}={6}\)

\(\displaystyle{4}{\left({x}-{4}\right)}^{2}-{\left({y}+{2}\right)}^{2}={0}\)

\(\displaystyle{\left[{2}{\left({x}-{4}\right)}\right]}^{2}={\left({y}+{2}\right)}^{2}\)

\(\displaystyle{2}{x}-{8}={\left({y}+{2}\right)}{\quad\text{or}\quad}{2}{x}-{8}=-{\left({y}+{2}\right)}\)

\(\displaystyle{2}{x}-{8}={y}+{2}{\quad\text{or}\quad}{2}{x}-{8}=-{y}-{2}\)

\(\displaystyle{2}{x}-{y}={10}{\quad\text{or}\quad}{2}{x}+{y}={6}\)

Step 3

Answer:

The equation represents pair of lines crossing each other at (4, -2) and the equations of the lines are \(\displaystyle{2}{x}–{y}={10}{\quad\text{and}\quad}{2}{x}+{y}={6}\)