Question 1009238
let a positive number be {{{x}}}
if the {{{square}}} of a positive number increased by {{{4}}} times the number is equal to {{{140}}}, we have

 {{{x^2+4x=140}}}

{{{x^2+4x-140=0}}}

{{{x^2+4x-140=0}}}

{{{x^2+14x-10x-140=0}}}

{{{(x^2+14x)-(10x+140)=0}}}

{{{x(x+14)-10(x+14)=0}}}

{{{(x-10)(x+14) = 0}}}

solutions:

if {{{(x-10) = 0}}}=>{{{x=10}}}
if {{{(x+14) = 0}}}=>{{{x=-14}}}

so, the number could be {{{10}}} or {{{-14}}}

but, since given condition is that we need "a positive number", your solution is number {{{highlight(10)}}}