Question 642216
{{{x^2-8x-4=0}}}
{{{x^2-8x = 4}}}
take half 'b', square it, and add to both sides:
{{{x^2-8x+16 = 4+16}}}
{{{(x-4)(x-4) = 20}}}
{{{(x-4)^2 = 20}}}
take the square root of both sides:
{{{(x-4) = sqrt(20)}}}
{{{x-4 = sqrt(2*2*5)}}}
{{{x-4 = 2*sqrt(5)}}}
{{{x = 2*sqrt(5)+4}}}
so x can be two possibilities:
{{{x = 2*sqrt(5)+4}}}
and
{{{x = -2*sqrt(5)+4}}}
.
or approximately,
x = {-0.47, 8.47}