Given the equation;
[tex]4x^2+5y^2+y+1=0[/tex]
We shall begin by Subtracting 5y^2 + y from both sides;
[tex]\begin{gathered} 4x^2+5y^2+y+1-5y^2-y=0-5y^2-y \\ 4x^2+1=-5y^2-y \\ \text{Factor out -1 from the right hand side;} \\ 4x^2+1=-1(5y^2+y) \end{gathered}[/tex]
Next step we subtract 1 from both sides;
[tex]\begin{gathered} 4x^2+1-1=-1(5y^2+y)-1 \\ 4x^2=-(5y^2+y)-1 \\ \end{gathered}[/tex]
Next step we take the square root of both sides;
[tex]\begin{gathered} \sqrt[]{4x^2}=\pm\sqrt[]{-(5y^2+y)-1} \\ 2x=\pm\sqrt[]{-(5y^2+y)-1} \end{gathered}[/tex]
We can now open the parenthesis on the right hand side;
[tex]\begin{gathered} 2x=\pm\sqrt[]{-5y^2-y-1} \\ \text{Divide both sides by 2;} \\ x=\frac{\pm\sqrt[]{-5y^2-y-1}}{2} \end{gathered}[/tex][tex]undefined[/tex]
