Respuesta :

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[tex]ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ \text{then exist two solutions}\\\\\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ \text{then exist one solution}\\\\x_o=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ \text{then not exist any solutions}[/tex]

[tex]x^2+13x+30=0\\\\a=1;\ b=13;\ c=30\\\\\Delta=13^2-4\cdot1\cdot30=169-120=49 > 0\\\\\sqrt\Delta=\sqrt{49}=7\\\\x_1=\dfrac{-13-7}{2\cdot1}=\dfrac{-20}{2}=-10\\\\x_2=\dfrac{-13+7}{2\cdot1}=\dfrac{-6}{2}=-3[/tex]

Answer: x = -10 or x= -6.

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