Question 880174
The left side is a perfect square already, as is too the right side.  You would take {{{(8/2)^2=4^2=16}}} which is already present on the left side member.


{{{(x+4)^2=36}}}
{{{x+4=0+- sqrt(36)}}}
{{{x+4=0+- 6}}}
{{{x=-4+- 6}}}
-
x is 2 or -10.


You can manage the given quadratic equation as
{{{x^2+8x+16-36=0}}}
{{{x^2+8x-20=0}}}
and you can identify your c from {{{ax^2+bx+c=0}}} as {{{c=-20}}}.