Question 993048
Let the numbers be x and y
{{{x-y = 5}}} so {{{x = y+5}}}
{{{x*y = 14}}}
{{{(y+5)*y = 14}}}
{{{y^2 +5y = 14}}}
{{{y^2 + 5y - 14 = 0}}}
{{{(y+7)(y-2)=0}}}
y = -7 and y=2 are possible solutions
if y = -7, then x = -2
if y = 2 then x = 7

so there are two sets of solution (-2, -7) and (7,2)