Question

Fórmula de Bhaskara

$3 { \times }^{2} - 15 \times + 12 = 0$
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Answer 1

$\tt{\purple{Vamos~l\acute{a}!!!}}$

Para resolvermos essa $\sf{\green{equac_{\!\!\!,}\tilde{a}o~do~2^{o}~grau}}$ precisamos de descobrir o valor do $\sf{\green{delta}}$ sabendo que resolvermos o delta com essa formula $\Rightarrow\sf{\green{~\Delta=b^{2}-4ac}}$ após descobrirmos o valor do delta precisamos descobrir o valor do $\sf{\tt{x}}$ que damos essa formula $\Rightarrow\sf{x=\green{~\dfrac{-b\pm\sqrt{\Delta} }{2a} }}$ após isso descobrirmos os dois valores do x.

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Lembre-se:

$\boxed{\begin{array}{lr}\sf{\red{a}x^{2}+\red{b}x+\red{c}=0}\end{array}}$

Ent:

$\boxed{\begin{array}{lr}\sf{\green{a}=\red{~~3}}\\\\\sf{\green{b}=\red{-15}}\\\\\sf{\green{c}=\red{12}}\end{array}}$

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$\tt{\pink{Resposta:}}$

$\boxed{~\tt{\red{S=4,1}}~}$

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$\tt{\blue{Calculo:}}$

$\boxed{\begin{array}{lr}\green{\tt{3x^{2}-15x+12=0 }}\end{array}}$

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Delta:

$\boxed{\begin{array}{lr}\sf{\Delta\green{~=(-15)^{2}-4.~3.~12}} \\\\\sf{~\Delta=225-144}\\\\\sf{\Delta=\red{81}}\end{array}}$

x :

$\boxed{\begin{array}{lr}\sf{x=\green{~\dfrac{-b\pm\sqrt{\Delta} }{2a} }}\\\\\\\sf{x=\dfrac{15\pm\sqrt{81} }{2.3} }\\\\\\\sf{x=\dfrac{15\pm9 }{6}}\end{array}}$

 $\tt{\green{x^{1}}=\red{4}}\\\\\tt{\green{x^{2}}=\red{1}}$

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$\Huge{\rm{\diagup\!\!\!\!A\diagup\!\!\!\!s\diagup\!\!\!\!s:\color{lime}\diagup\!\!\!\!T~~\diagup\!\!\!\!O~~\diagup\!\!\!\!D~~\diagup\!\!\!\!D~~\diagup\!\!\!\!Y~~}}\maltese$

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