C-05-Q025mediumsingle_mcqThe roots of 3x2+7x+1=03x^2 + 7x + 1 = 03x2+7x+1=0 are —a−7±376\dfrac{-7 \pm \sqrt{37}}{6}6−7±37b7±376\dfrac{7 \pm \sqrt{37}}{6}67±37c−7±496\dfrac{-7 \pm \sqrt{49}}{6}6−7±49d−7±136\dfrac{-7 \pm \sqrt{13}}{6}6−7±13