[tex]\displaystyle x^2-7x+10=0\\\\x = \frac{-b \pm \sqrt{\Delta}}{2a}\\\\x = \frac{-(-7) \pm \sqrt{b^2-4ac}}{2.1}\\\\x = \frac{7 \pm \sqrt{(-7)^2-4.1.10}}{2.1}\\\\x = \frac{7 \pm \sqrt{9}}{2}\\\\x = \frac{7 \pm 3}{2}\\\\\\x' = \frac{7+3}{2} =\frac{10}{2} = 5\\\\x'' = \frac{7-3}{2} = \frac{4}{2} = 2[/tex]
Com isso, os catetos medem 5 cm e 2 cm, agora é só calcular a hipotenusa.
[tex]h^2 = (x')^2 + (x'')^2\\h^2 = 5^2 + 2^2\\h^2 = 25 + 4\\h^2 = 29\\h = \sqrt{29} \mbox{ } cm[/tex]
