Resposta :
Temos que:
[tex]\text{t}^2-6\text{t}+5=0[/tex]
[tex]\text{t}=\dfrac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot5}}{2\cdot1}=\dfrac{6\pm4}{2}[/tex]
Logo, as raízes são:
[tex]\text{t}'=\dfrac{6+4}{2}=5[/tex]
[tex]\text{t}"=\dfrac{6-4}{2}=1[/tex]
[tex]t^2-6t+5=0\\\Delta=b^2-4*a*c\\\Delta=(-6)^2-4*1*5\\\Delta=36-20\\\Delta=16\\\\[/tex] agora calcula-se bhaskara:
[tex]x=\frac{-b^+_-\sqrt\Delta}{1*a}\\x=\frac{6^+_-\sqrt16}{2*1}\\x=\frac{6^+_-4}{2}\\\\x_1=6+4/2\\x_1=10/2\\\x_1=5\\\\x_2=6-4/2\\x_2=2/2\\x_2=1\\\\S= (5,1)[/tex]