Question 126363
You have 2 problems with complex number division and 4 quadratic equations.  I'll do one of each:


#1:  To do complex number division, multiply both numerator and denominator by the conjugate of the complex number.  The conjugate is formed by changing the sign on the imaginary term.


{{{(1+i)/(3-2i)}}}.  The conjugate of {{{3-2i}}} is {{{3+2i}}}, so:


{{{((1+i)(3+2i))/((3-2i)(3+2i))}}}.


Use FOIL on the numerator and the factorization of the difference of two squares on the denominator, simplify the expression.  Remember that {{{i^2=-1}}}, 


{{{(3+2i+3i-2)/(9+4)}}}


{{{(1+5i)/13}}}.   Done.


#4. {{{x^2+2x+5=0}}}


Use the quadradic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


{{{x = (-2 +- sqrt( 2^2-4*1*5 ))/(2*1) }}} 


{{{x = (-2 +- sqrt( 4-20 ))/(2) }}} 


{{{x = (-2 +- sqrt( -16 ))/(2) }}}


{{{x[1]=-1+2i}}}, {{{x[2]=-1-2i}}}