Question 682033
I'll do the first one to get you started (Please only post one problem at a time. Thanks)


Use the quadratic formula to solve for m


{{{m = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{m = (-(6)+-sqrt((6)^2-4(2)(2)))/(2(2))}}} Plug in {{{a = 2}}}, {{{b = 6}}}, {{{c = 2}}}


{{{m = (-6+-sqrt(36-(16)))/(4)}}}


{{{m = (-6+-sqrt(20))/4}}}


{{{m = (-6+sqrt(20))/4}}} or {{{m = (-6-sqrt(20))/4}}}


{{{m = (-6+2*sqrt(5))/4}}} or {{{m = (-6-2*sqrt(5))/4}}}


{{{m = (-3+sqrt(5))/2}}} or {{{m = (-3-sqrt(5))/2}}}


{{{m = -0.381966}}} or {{{m = -2.618034}}}


{{{m = -0.382}}} or {{{m = -2.618}}}


So the approximate solutions to {{{2m^2 + 6m + 2 = 0}}} are {{{m = -0.382}}} or {{{m = -2.618}}}