Resposta:
[tex]x'=-3\\\\x''=2[/tex]
Explicação passo a passo:
[tex]f(x)=x^2+x-6[/tex]
onde:
[tex]a=1;\\b=1;\\c=-6;[/tex]
Pela fórmula de Bháskara:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-6)}}{2\cdot1}\\\\x=\frac{-1\pm\sqrt{1+24}}{2}\\\\x=\frac{-1\pm\sqrt{25}}{2}\\\\x=\frac{-1\pm5}{2}\\\\x'=\frac{-1-5}{2}=\frac{-6}{2}=-3\\\\x''=\frac{-1+5}{2}=\frac{4}{2}=2[/tex]
