Question 623213
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Hi,
18z + 45 + z^2   \Note use of ^  (uppercase 6)
If You meant:
f(z) = {{{z^2 + 18z + 45}}}   and to find where f(z) = 0
{{{z^2 + 18z + 45}}} = 0
Quadratic formula ALWAYS works
{{{z = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{z = (-18 +- sqrt( 18^2-4*1*45 ))/(2) }}}
{{{z = (-18 +- sqrt(144))/(2) }}}
{{{z = (-18 +- 12)/(2) }}}
     z = -3  and z = -15
OR by factoring:
{{{z^2 + 18z + 45}}} = 0
  (z + 3)(z+15) = 0