Respuesta :

[tex]x^2=-5x-3\\x^2+5x+3=0\\\\a=1;\ b=5;\ c=3\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot3=25-12=13 \ \textgreater \ 0\\\\then\\x_1=\dfrac{-b-\sqer\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-5-\sqrt{13}}{2\cdot1}=\boxed{\dfrac{-5-\sqrt{13}}{2}}\\\\x_2=\dfrac{-5+\sqrt{13}}{2\cdot1}=\boxed{\dfrac{-5+\sqrt{13}}{2}}[/tex]