Resposta :
Resposta:
[tex]\sf \displaystyle x^{2} - x - 6 = 0[/tex]
[tex]\sf \displaystyle ax^{2} +bx + c = 0[/tex]
a = 1
b = - 1
c = - 6
Determinar o Δ:
[tex]\sf \displaystyle \Delta = b^2 -\:4ac[/tex]
[tex]\sf \displaystyle \Delta = (-1)^2 -\:4 \cdot 1 \cdot (-6)[/tex]
[tex]\sf \displaystyle \Delta = 1 + 24[/tex]
[tex]\sf \displaystyle \Delta = 25[/tex]
Determinar as raízes da equação:
[tex]\sf \displaystyle \dfrac{-\,b \pm \sqrt{ \Delta } }{2a} = \dfrac{1 \pm \sqrt{ 25 } }{2\cdot 1} = \dfrac{1 \pm 5 }{2} \Rightarrow\begin{cases} \sf x_1 = &\sf \dfrac{1 + 5}{2} = \dfrac{6}{2} = \;3 \\\\ \sf x_2 = &\sf \dfrac{1 - 5}{2} = \dfrac{- 4}{2} = - 2\end{cases}[/tex]
[tex]\sf \boldsymbol{ \sf \displaystyle S = \{ x \in \mathbb{R} \mid x = -\: 3 \mbox{\sf \;e } x = 2 \} }[/tex]
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