Question 1124445

if a negative number is {{{-x}}}

and if the square of a negative number is {{{54}}} more than {{{three}}} times the negative number, we have

{{{(-x)^2=3*(-x)+54}}}...since {{{(-x)^2=(-x)(-x)=x^2}}},we have

{{{x^2=-3x+54}}}

{{{x^2+3x-54=0}}}...actor

{{{x^2+9x-6x-54=0}}}

{{{(x^2+9x)-(6x+54)=0}}}

{{{x(x+9)-6(x+9)=0}}}

{{{(x - 6) (x + 9) = 0}}}

solutions:

if {{{(x - 6)  = 0}}}->{{{x=6}}}-> since we are looking for negative number, disregard this solution

i {{{ (x + 9) = 0}}}->{{{highlight(x=-9)}}}-> your number