What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

Answer:   [tex]x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}[/tex]Step-by-step explanation:My favorite way to go at this is to look at a graph. It shows the vertex at (-2, -11). Since the leading coefficient is 2, this means the roots are ...   [tex]-2\pm\sqrt{\dfrac{11}{2}}[/tex]where the 2 in the denominator of the radical is the leading coefficient.__You can also use other clues:the axis of symmetry is -b/(2a) = -8/(2(2)) = -2, so answer choices C and D don't workthe single change in sign in the coefficients (+ + -) tells you there is one positive real root, so answer choice B doesn't work.The first answer choice is the only one with values symmetrical about -2 and one of them positive.__You may be expected to use the quadratic formula:   [tex]x=\dfrac{-b\pm\sqrt{b^-4ac}}{2a}=\dfrac{-8\pm\sqrt{8^2-4(2)(-3)}}{2(2)}\\\\=\dfrac{-8}{4}\pm\dfrac{\sqrt{88}}{4}=-2\pm\sqrt{\dfrac{11}{2}}[/tex]