Question 913506
x = first number


x+1 = second number



{{{x*(x+1) = 20}}}


{{{x^2+x = 20}}}


{{{x^2+x-20=0}}}


Use the quadratic formula to solve for x


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


{{{x = (-(1)+-sqrt((1)^2-4(1)(-20)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = 1}}}, {{{c = -20}}}  


{{{x = (-1+-sqrt(1-(-80)))/(2)}}}


{{{x = (-1+-sqrt(1+80))/(2)}}}


{{{x = (-1+-sqrt(81))/2}}}


{{{x = (-1+sqrt(81))/2}}} or {{{x = (-1-sqrt(81))/2}}}


{{{x = (-1+9)/2}}} or {{{x = (-1-9)/2}}}


{{{x = 8/2}}} or {{{x = -10/2}}}


{{{x = 4}}}    or    {{{x = -5}}}


If you're only considering positive numbers, the two numbers are 4 and 5


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Thanks,


Jim