As raízes da equação de 2º grau são: x'= -1 e x"= -(5/2).
Acompanhe a solução:
Aplicando a fórmula de Bháskara, temos?
[tex]2x^2+7x+5=0\\\\\\\Delta=b^2-4ac\\\\\Delta=7^2-4\cdot2\cdot 5\\\\\Delta=49-40\\\\\boxed{\Delta=9}\\\\\\x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\dfrac{-7\pm\sqrt{9}}{2\cdot2}\\\\\boxed{x=\dfrac{-7\pm3}{4}}[/tex]
[tex]\boxed{\large\begin {array}{l}x'=\dfrac{-7+3}{4}\\\\x'=\dfrac{-4}{4}\\\\\Large\boxed{\boxed{x'=-1}}\Huge\checkmark\end {array}} \quad\quad \boxed{\large\begin {array}{l}x"=\dfrac{-7-3}{4}\\\\x"=\dfrac{-10}{4}\\\\\Large\boxed{\boxed{x"=-\dfrac{5}{2}}}\Huge\checkmark\end {array}}[/tex]
Resposta:
Portanto, as raízes da equação de 2º grau é x'= -1 e x"= -(5/2).
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