Question 1137906

let a positive number be {{{x}}}

if the square of a positive number is decrease by five times the number, the result is {{{14}}}, then we have

{{{x^2-5x=14}}}....solve for {{{x}}}

{{{x^2-5x-14=0}}}

{{{x^2+2x-7x-14=0}}}

{{{(x^2+2x)-(7x+14)=0}}}

{{{x(x+2)-7(x+2)=0}}}

{{{(x-7)(x+2)=0}}}

solutions:

{{{x=7}}}
or
{{{x=-2}}}-> since you need positive number, disregard negative solution

so, your number is {{{7}}}