The given triangle in the question is a right-angled triangle. In order to get the length of each side, we will apply the Pythagoras theorem.
The Pythagoras theorem is,
[tex]\text{Hypotenuse}^2=Opposite^2+Adjacent^2[/tex]Where,
[tex]\begin{gathered} \text{Hypotenuse}=5 \\ \text{Opposite}=2x-2 \\ \text{Adjacent}=x \end{gathered}[/tex]Therefore,
[tex]5^2=(2x-2)^2+x^2[/tex]Let us expand the above
[tex]\begin{gathered} 25=2x(2x-2)-2(2x-2)+x^2 \\ 25=4x^2-4x-4x+4+x^2 \\ 25=5x^2-8x+4 \end{gathered}[/tex]Switch sides
[tex]5x^2-8x+4=25[/tex]Subtract 25 from both sides
[tex]5x^2-8x+4-25=25-25[/tex]Simplify
[tex]5x^2-8x-21=0[/tex]Solve with the quadratic formula
[tex]x_{1,\: 2}=\frac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4\cdot\:5\left(-21\right)}}{2\cdot\:5}[/tex]Thus
[tex]\sqrt[]{(-8)^2-4\cdot\: 5(-21)}=22[/tex]Therefore,
[tex]x_{1,\: 2}=\frac{-\left(-8\right)\pm\:22}{2\cdot\:5}[/tex]Separate the solutions
[tex]x_1=\frac{-\left(-8\right)+22}{2\cdot\:5},\: x_2=\frac{-\left(-8\right)-22}{2\cdot\:5}[/tex]Hence,
[tex]\begin{gathered} x=\frac{-(-8)+22}{2\cdot\: 5}=3 \\ x=\frac{-(-8)-22}{2\cdot\: 5}=-\frac{7}{5} \end{gathered}[/tex]The solutions to the quadratic equations are
[tex]x=3,\: x=-\frac{7}{5}[/tex]Therefore, from the above result, the length of a triangle can never be negative.
Hence, x = 3
Let us now solve the length of the remaining side
[tex]\begin{gathered} 2x-2=2(3)-2=6-2=4 \\ \therefore2x-2=4 \end{gathered}[/tex]Therefore, the length of each leg is
[tex]3ft,4ft,5ft[/tex]