If $x,y\in R$  and $2x^2+6xy+5y^2=1,$  then

   Given equation is $2x^2+6xy+5y^2=1$ -----(1)

     Eq. (1) can be rewritten as $2x^2+\left(6y\right)x+5y^2-1=0$

              Since $x$  is real,

            $36y^2-8\left(5y^2-1\right)\ge 0$

$\left(1\right)$        $\Rightarrow y^2\le 2$

            $\Rightarrow -\sqrt{2}\le y\le \sqrt{2}$

           Eq (1) can also be re written as

            $5y^2+\left(6x\right)y+2x^2-1=0$

$\left(2\right)$      Since$y$  is real,

         $36x^2-20\left(2x^2-1\right)\ge 0$

         $\Rightarrow 36x^2-40x^2+20\ge 0$

          $\Rightarrow -4x^2\ge -20$

          $\Rightarrow x^2\le 5$

         $\Rightarrow -\sqrt{5}\le x\le \sqrt{5}$