Resposta :
Por bhaskara
[tex]6x^2+x-1=0\\\Delta=1-4.6(-1)\\\Delta=25[/tex]
[tex]x= \frac{-1\pm5}{12} \left \{ {{x'= \frac{-1+5}{12}= \frac{4}{12}= \frac{1}{3} } \atop {x''= \frac{-1-5}{12}= \frac{-1}{2} }} \right. [/tex]
S={1/3 ,-1/2}
[tex]6x^2+x-1=0\\\Delta=1-4.6(-1)\\\Delta=25[/tex]
[tex]x= \frac{-1\pm5}{12} \left \{ {{x'= \frac{-1+5}{12}= \frac{4}{12}= \frac{1}{3} } \atop {x''= \frac{-1-5}{12}= \frac{-1}{2} }} \right. [/tex]
S={1/3 ,-1/2}