[tex]\begin{gathered} longer=x+7 \\ shorter=x \\ hypotenuse=13 \\ x^2+(x+7)^2=13^2 \\ x^2+x^2+2(x)(7)+7^2=13^2 \\ 2x^2+14x+49=169 \\ 2x^2+14x+49-169=0 \\ 2x^2+14x-120=0 \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=2 \\ b=14 \\ c=-120 \\ x=\frac{-(14)\pm\sqrt{(14)^2-4(2)(-120)}}{2(2)} \\ x=\frac{-14\pm\sqrt{196+960}}{4} \\ x=\frac{-14\pm\sqrt{1156}}{4} \\ x=\frac{-14\pm34}{4} \\ x1=\frac{-14+34}{4}=\frac{20}{4}=5 \\ x1=5 \\ x2=\frac{-14-34}{4}=\frac{-48}{4}=-12,\text{ this is not the answer because the length} \\ is\text{ not a negative number} \\ Hence \\ longer=x+7 \\ longer=5+7 \\ longer=12 \\ shorter=5 \\ Therefore,\text{ the lengths of the triangle are 5 and 12} \end{gathered}[/tex]
