Question 61508
Yes, {{{sqrt(-1)=i}}}.
:
x(x-4)=-8
{{{x^2-4x=-8}}}
{{{x^2-4x+8=-8+8}}}
{{{x^2-4x+8=0}}}
a=1, b=-4, and c=8
The quadratic formula is:
{{{highlight(x=(-b+-sqrt(b^2-4ac))/(2a))}}}
{{{x=(-(-4)+-sqrt((-4)^2-4(1)(8)))/(2(1))}}}
{{{x=(4+-sqrt(16-32))/2}}}
{{{x=(4+-sqrt(-16))/2}}}
{{{x=(4-4i)/2}}} and {{{x=(4+4i)/2}}}
{{{x=4/2-4i/2}}} and {{{x=4/2+4i/2}}}
{{{x=2-2i}}} and {{{x=2+2i}}}
Happy Calculating!!!