Step 1

Let:

\(\displaystyle{x}=\text{distance of the ladder from the bottom of the wall}\)

\(\displaystyle{y}=\text{height of the ladder from the bottom of the wall}\)

\(\displaystyle{L}=\text{length of the ladder}\)

Using Pythagorean theorem

\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\)

Since the height is 3 feet farther up than the distance:

\(\displaystyle{y}={x}+{3}\)

\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{\left({x}+{3}\right)}^{{{2}}}\)

\(\displaystyle{\left({15}\right)}^{{{2}}}={x}^{{{2}}}+{x}^{{{2}}}+{6}{x}+{9}\)

\(\displaystyle{225}={2}{x}^{{{2}}}+{6}{x}+{9}\)

\(\displaystyle{2}{x}^{{{2}}}+{6}{x}-{216}={0}\)

\(\displaystyle{\left({x}-{9}\right)}{\left({x}+{12}\right)}={0}\)

\(\displaystyle{x}{1}={9}\) and \(\displaystyle{x}{2}=-{12}\) (neglect the negative value)

Answer: \(\displaystyle{x}={9}\text{feet}\)

Let:

\(\displaystyle{x}=\text{distance of the ladder from the bottom of the wall}\)

\(\displaystyle{y}=\text{height of the ladder from the bottom of the wall}\)

\(\displaystyle{L}=\text{length of the ladder}\)

Using Pythagorean theorem

\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\)

Since the height is 3 feet farther up than the distance:

\(\displaystyle{y}={x}+{3}\)

\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{\left({x}+{3}\right)}^{{{2}}}\)

\(\displaystyle{\left({15}\right)}^{{{2}}}={x}^{{{2}}}+{x}^{{{2}}}+{6}{x}+{9}\)

\(\displaystyle{225}={2}{x}^{{{2}}}+{6}{x}+{9}\)

\(\displaystyle{2}{x}^{{{2}}}+{6}{x}-{216}={0}\)

\(\displaystyle{\left({x}-{9}\right)}{\left({x}+{12}\right)}={0}\)

\(\displaystyle{x}{1}={9}\) and \(\displaystyle{x}{2}=-{12}\) (neglect the negative value)

Answer: \(\displaystyle{x}={9}\text{feet}\)