with the pythagorean theorem
[tex]\begin{gathered} (4+3x)^2=x^2+(3+3x)^2 \\ 16+24x+9x^2=x^2+9+18x+9x^2 \\ 16+24x+9x^2=10x^2+18x+9 \\ 16+24x+9x^2-9=10x^2+18x+9-9 \\ 9x^2+24x+7=10x^2+18x \\ 9x^2+24x+7-18x=10x^2+18x-18x \\ 9x^2+6x+7=10x^2 \\ 9x^2+6x+7-10x^2=10x^2-10x^2 \\ -x^2+6x+7=0 \end{gathered}[/tex]
using the formula of the quadratic equation
[tex]\begin{gathered} x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x1=\frac{-6+\sqrt{6^2-4\left(-1\right)\cdot\:7}}{2\left(-1\right)}=-1 \\ x2=\frac{-6-\sqrt{6^2-4\left(-1\right)\cdot\:7}}{2\left(-1\right)}=7 \end{gathered}[/tex]
the length cannot be negative, therefore x=7
length of the shorter leg is: 7ft
length of the longer leg is: 3+3(7)= 24ft
length of the hypotenuse is: 4+3(7)= 25ft
