Question 174686
I like to use the method called "completing the square"
To do this you first get the constant term on the 
other side
(1) Add {{{80}}} to both sides
{{{x^2 + 11x - 80 = 0}}}
{{{x^2 + 11x = 80}}}
(2) Now divide the coefficient of the x term by {{{2}}},
then square it and add this to both sides
{{{x^2 + 11x + (11/2)^2 = 80 + (11/2)^2}}}
The purpose of this is to make the left side a perfect
square, which it is. It might be hard to see, but I
can rewrite this as
{{{(x + 11/2)^2 = 80 + (11/2)^2}}}
Now I want to express {{{80}}} as {{{320/4}}}
{{{(x + 11/2)^2 = 320/4 + 121/4}}}
{{{(x + 11/2)^2 = 441/4}}}
Now take the square root of both sides
{{{x + 11/2 = 0 +-sqrt(441)/2}}}
{{{x = (-11 + 21)/2}}}
{{{x = 5}}}
and
{{{x = (-11 - 21)/2}}}
{{{x = -16}}}
check answers:
{{{x^2 + 11x - 80 = 0}}}
{{{5^2 + 11*5 - 80 = 0}}}
{{{25 + 55 - 80 = 0}}}
{{{0 = 0}}}
OK
{{{(-16)^2 + 11*(-16) - 80 = 0}}}
{{{256 - 176 - 80 = 0}}}
{{{256 - 256 = 0}}}
{{{0 = 0}}}
OK