C-05-Q004mediumsingle_mcqThe two roots of ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 are given by —ax=−b±b2+4ac2ax = \dfrac{-b \pm \sqrt{b^2 + 4ac}}{2a}x=2a−b±b2+4acbx=b±b2−4ac2ax = \dfrac{b \pm \sqrt{b^2 - 4ac}}{2a}x=2ab±b2−4accx=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4acdx=−b±4ac−b22ax = \dfrac{-b \pm \sqrt{4ac - b^2}}{2a}x=2a−b±4ac−b2