Answer:
The solution to the quadratic equation is;
[tex]\begin{gathered} y=-1+\frac{\sqrt[]{2}}{2} \\ \text{and} \\ y=-1-\frac{\sqrt[]{2}}{2} \\ y=-1+\frac{\sqrt[]{2}}{2},-1-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]
Explanation:
Given the quadratic equation;
[tex]2y^2+4y+1=0[/tex]
Applying quadratic formula;
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Substituting the coefficients of the quadratic equation;
[tex]\begin{gathered} y=\frac{-4\pm\sqrt[]{4^2-4(2)(1)}}{2(2)} \\ y=\frac{-4\pm\sqrt[]{16^{}-8}}{4} \\ y=\frac{-4\pm\sqrt[]{8}}{4} \\ y=\frac{-4\pm2\sqrt[]{2}}{4} \\ y=\frac{-2\pm\sqrt[]{2}}{2} \end{gathered}[/tex]
Therefore, the solution to the quadratic equation is;
[tex]\begin{gathered} y=-1+\frac{\sqrt[]{2}}{2} \\ \text{and} \\ y=-1-\frac{\sqrt[]{2}}{2} \\ y=-1+\frac{\sqrt[]{2}}{2},-1-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]
