Resposta :

3478elc
 x^2 -5x + 6 = 0

delta = (-5)^2 - 4.1.6 = 25 - 24 = 1

x=5 +/-V1 ==> x= 5 +/- 1 ==> x1=5 + 1==>x1= 3 ; x2= 5 -1 ==> x2= 2
       2.1                   2                      2                             2
ArthurPDC
Resolvendo a equação, temos:


[tex]x^2-5x+6=0\\\\ \Delta=b^2-4\cdot a\cdot c\\ \Delta=(-5)^2-4\cdot1\cdot6\\ \Delta=25-24\\ \Delta=1\\\\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{5\pm\sqrt{1}}{2\cdot1}=\dfrac{5\pm1}{2}\Longrightarrow\begin{cases}x_1=\frac{5+1}{2}=\frac{6}{2}=3\\x_2=\frac{5-1}{2}=\frac{4}{2}=2\end{cases}\\\\ S=\{2,3\}[/tex]

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