Question

6(2x+5)=-3(-x+2)x por favor
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Answer 1

El ejercicio a resolver es:

$\huge \boxed{\sf{6(2x+5)=-3(-x+2)x}}$

↓

Ahora procedemos a desarrollar:

$\large\displaystyle\begin{gathered}\sf 6(2x+5)=(-3(-x+2))(x) \end{gathered}$

$\large\displaystyle\begin{gathered}\sf 12x+30=3x^{2} -6x \ \ \to \ \ \ Simplificar \ ambos \ lados \ de \ la \ ecuacion. \end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\sf 12x+30-(\blue{3x^{2} -6x)}=3x^{2} -6x-(\blue{3x^{2} -6x}) \ \to \ Restar \ 3x^{2} -6 \ de \ ambos \ lados \end{gathered} }$

$\large\displaystyle\begin{gathered}\sf -3x^{2} +18x+30=0 \end{gathered}$

$\large\displaystyle\begin{gathered}\sf Para \ esta \ ecuacion:\blue{a=-3},\green{b=18},\red{c=30} \end{gathered}$

$\large\displaystyle\begin{gathered}\sf \blue{-3}x^{2} +\green{18}x+\red{30}=0 \end{gathered}$

$\large\displaystyle\begin{gathered}\sf Usar \ la \ formula \ cuadratica \ con:\blue{a=-3},\green{b=18},\red{c=30} \end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\sf x=\dfrac{-\green{B}\pm\sqrt{\green{B}^{2}-4\blue{a}\red{C} } }{2\blue{a}} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\sf x=\dfrac{-(\green{18})^{2}\pm\sqrt{(\green{18})^{2}-4(\blue{-3})(\red{30)} } }{2(\blue{-3})} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\sf x=\dfrac{-18\pm\sqrt{684} }{-6} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\sf \red{x=3-\sqrt{19} \ o \ x=3+\sqrt{19} } \end{gathered} }$

$\large\displaystyle\begin{gathered}\sf \underline{Respuesta:} \end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\sf \red{x=3\pm\sqrt{19} } \end{gathered} }$

Saludos Estivie