Question 1193021
<pre>
Another way is to solve f(x) = 0 for x

{{{2(x+1)^2-6=0}}}

{{{2(x+1)^2=6}}}

{{{(x+1)^2=3}}}

{{{x+1="" +- sqrt(3)}}}

{{{x=-1 +- sqrt(3)}}}

One zero if you use the +, and another
if you use the -.  That makes 2 zeros.

Edwin</pre>