Question 8591
I'm afraid in both exercises, you have tried to solve by completing the square on the z and t terms, but completing the square requires that the coefficient in the squared term MUST BE 1.  So completing the square in problems like this really gets ugly.  You are better off to use the quadratic formula:


1.  {{{5z^2 -4z+1 = 0}}}
a=5, b=-4, c=1
{{{z= (4+-sqrt(16-4*5*1))/(2*5)}}}
{{{z= (4 +-sqrt(-4))/10}}}
{{{z= (4+-i*2)/10}}}
{{{z= (2 +-i)/5}}}


2.  {{{5t^2 + 4t - 3 = 0}}}
a=5, b=4, c=-3

The result is similar.


R^2 at SCC