Question 463583


let number be {{{x}}}

if the {{{sum}}} of a {{{number}}} and {{{its}}}{{{ square}}} is {{{20}}}, you have

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

{{{x^2+x-20=0}}}..use quadratic formula

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

{{{x = (-1 +- sqrt( 1^2-4*1*(-20) ))/(2*1) }}}

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

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

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


solutions:

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

{{{x = 8/2 }}}

{{{x = 4 }}}

or

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

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

{{{x =-5 }}}


check:

if {{{x = 4 }}}:{{{x+x^2=20}}}...-> {{{4+4^2=20}}}.-> {{{4+16=20}}}.-> {{{20=20}}}


if {{{x = -5 }}}:{{{-5+(-5)^2=20}}}...-> {{{-5+25=20}}}.-> {{{20=20}}}