Difference of legs

In a right triangle, the length of the hypotenuse is 65 m, and the difference of legs is 23 m. Calculate the perimeter of this triangle.

Correct result:

o =  154 m

Solution:

$$\begin{gathered} c=65\text{ m } \\ a=b+23\text{ m } \\ {c}^2=a^2+b^2 \\ 65^2=(b+23{)}^2+b^2 \\ 65^2=(b+23{)}^2+b^2 \\ -2b^2-46b+3696=0 \\ 2b^2+46b-3696=0 \\ p=2;q=46;r=-3696 \\ D=q^2-4pr=46^2-4\cdot 2\cdot (-3696)=31684 \\ D>0 \\ {b}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{-46\pm \sqrt{31684}}{4}} \\ {b}_{1,2}={\displaystyle \frac{-46\pm 178}{4}} \\ {b}_{1,2}=-11.5\pm 44.5 \\ {b}_1=33 \\ {b}_2=-56 \\ \text{ Factored form of the equation: } \\ 2(b-33)(b+56)=0 \\ b>0 \\ b=b_1=33=33\text{ m } \\ a=b+23=33+23=56\text{ m } \\ o=a+b+c=56+33+65=154\text{ m} \end{gathered}$$

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