Question 216313
Let n = the number, then:
{{{n+2n^2 = 6}}} Subtract 6 from both sides.
{{{2n^2+n-6 = 0}}} Factor.
{{{(2n-3)(n+2) = 0}}} Apply the zero product rule.
{{{2n-3 = 0}}} or {{{n+2 = 0}}} which means that...
{{{highlight(n = 3/2)}}} or {{{highlight(n = -2)}}}
Check:
{{{(3/2)+2(3/2)^2 = (3/2)+2(9/4)}}}={{{(3/2)+(9/2) = 12/2}}}={{{6}}}
{{{(-2)+2(-2)^2 = (-2)+2(4)}}} = {{{-2+8 = 6}}}