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Solve the equation using the quadratic formula. 2y^2 + 4y + 1 = 0

Solve the equation using the quadratic formula. 2y^2 + 4y + 1 = 0-example-1
User Brightbyte
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4 votes

Answer:

The solution to the quadratic equation is;


\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}

Step-by-step explanation:

Given the quadratic equation;


2y^2+4y+1=0

Applying quadratic formula;


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting the coefficients of the quadratic equation;


\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}

Therefore, the solution to the quadratic equation is;


\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}

User Nick Betcher
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