Question 642261
solve {{{x}}} for {{{y=0}}}  

{{{y=2x^2+ 1}}}......plug in {{{y=0}}} 

{{{0=2x^2+ 1}}}

{{{-1=2x^2}}}

{{{-1/2=x^2}}}

{{{sqrt(-1/2)=x}}}

solutions:

{{{sqrt(-1/2)=x}}}...=>..{{{sqrt(-1)/sqrt(2)=x}}}..=>...{{{i/sqrt(2)=x}}}

{{{-sqrt(-1/2)=x}}}..=>...{{{-i/sqrt(2)=x}}}