Resposta :
[tex](x - 1) . (x+3) = 5
\\x^{2}+3x-x-3=5\\\\x^{2}+2x-8=0\begin{cases}a=1\\b=2\\c=-8\end{cases}\\\\x=\frac{-b\±\sqrt{b^{2}-4ac}}{2a}\\\\x=\frac{-2\±\sqrt{2^{2}-4.1.(-8)}}{2.1}\\\\x=\frac{-2\±\sqrt{4+32}}{2}\\\\x=\frac{2\±6}{2}\begin{cases}x_{+}=\frac{8}{2}=4\\x_{-}=\frac{4}{2}=2\end{cases}[/tex]
(x - 1) . (x+3) = 5
x^2 + 3x - x - 3 - 5 = 0
x^2 + 2x - 8 = 0
delta= b^2 - 4ac
delta= 2^2-4.1.(-8) = 4 + 32= 36
x =-2 +/- V36 ==> x= -2 +/- 6
2.1 2
x1= -2+6 =>x1= 4 =>x1= 2 ; x2= -2-6 ==>x2= -8 =>x2= -4
2 2 2 2
x^2 + 3x - x - 3 - 5 = 0
x^2 + 2x - 8 = 0
delta= b^2 - 4ac
delta= 2^2-4.1.(-8) = 4 + 32= 36
x =-2 +/- V36 ==> x= -2 +/- 6
2.1 2
x1= -2+6 =>x1= 4 =>x1= 2 ; x2= -2-6 ==>x2= -8 =>x2= -4
2 2 2 2