Question 98234


{{{x^2+18x+80=0}}} Start with the given equation



{{{x^2+18x=-80}}} Subtract 80 from both sides



Take half of the x coefficient 18 to get 9 (ie {{{18/2=9}}})

Now square 9 to get 81 (ie {{{(9)^2=81}}})




{{{x^2+18x+81=-80+81}}} Add this result (81) to both sides. Now the expression {{{x^2+18x+81}}} is a perfect square trinomial.





{{{(x+9)^2=-80+81}}} Factor {{{x^2+18x+81}}} into {{{(x+9)^2}}}  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




{{{(x+9)^2=1}}} Combine like terms on the right side


{{{x+9=0+-sqrt(1)}}} Take the square root of both sides


{{{x=-9+-sqrt(1)}}} Subtract 9 from both sides to isolate x.


So the expression breaks down to

{{{x=-9+sqrt(1)}}} or {{{x=-9-sqrt(1)}}}



{{{x=-9+1}}} or {{{x=-9-1}}}    Take the square root of 1 to get 1



{{{x=-8}}} or {{{x=-10}}} Now combine like terms


So our answer is

{{{x=-8}}} or {{{x=-10}}}



Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+18x+80) }}} graph of {{{y=x^2+18x+80}}}


Here we can see that the x-intercepts are {{{x=-8}}} and {{{x=-10}}}, so this verifies our answer.