Question 927523


Twice a certain whole number {{{x}}}: {{{2x}}}

 subtracted from 3times the square of the number:{{{3x^2}}}  

leaves {{{133}}}:{{{3x^2-2x=133}}} 


{{{3x^2-2x=133}}}


{{{3x^2-2x-133=0}}}


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


{{{x = (-(-2) +- sqrt( (-2)^2-4*3*(-133) ))/(2*3) }}}


{{{x = (2 +- sqrt( 4+1596 ))/6 }}}

{{{x = (2 +- sqrt( 1600 ))/6 }}}


{{{x = (2 +- 40)/6 }}}

solutions:

{{{x = (2 + 40)/6 }}}

{{{x = 42/6 }}}

{{{x = 7 }}}

or

{{{x = (2 - 40)/6 }}}

{{{x =-38/6 }}}

{{{x = -6.33}}}


since we need a certain {{{whole}}} number, our solution is {{{highlight(x = 7) }}}