Question 1003671

If {{{5}}} times the {{{square}}} of a positive number {{{x}}} is added to the product of {{{31}}} and the number {{{x}}}, and the result is {{{28}}}, we have

{{{5x^2+31x=28}}}...........solve for {{{x}}}

{{{5x^2+31x-28=0}}}....factor completely

{{{5x^2-4x+35x-28=0}}}

{{{(5x^2-4x)+(35x-28)=0}}}

{{{x(5x-4)+7(x-4)=0}}}

{{{(x+7)(5x-4) = 0}}}

solutions:

if {{{(x+7) = 0}}}=>{{{x=-7}}}
if{{{(5x-4) = 0}}}=>{{{5x=4}}}=>{{{x=4/5}}}


Options are:
{{{4/5}}}=> your option
-4 
-7/5 
7