$$y^2-\frac25y=2$$

This is where I get:

$$\frac{\frac25 \pm\sqrt{(2/5)^2 - 4 \cdot 1 \cdot (-2)}}2$$ Within the square root part I get $8 +\frac4{25}$, I'm supposed to divide that by $2$. Supposedly I'm getting $\frac8{25}$ http://www.mathway.com/problem/NDcwMzE3OTIw

$$y=\frac{\dfrac25\pm\dfrac{2\sqrt{51}}{5}}{2\cdot 1}$$

Why did they get $\sqrt{51}$? I don't see how you can evaluate $(2/5)^2-4\cdot1\cdot(-2)$ and get that.

• I am getting $y_{0,1} = \frac{\frac{2}{5}\pm \frac{13}{5}}{2}$. That is $y_0=\frac{3}{2}, y_1 = -1$. – quapka Dec 9 '14 at 21:11
• @quapka That's wrong, the solutions are $$\frac15\left(1\pm\sqrt{51}\right)$$ – Alice Ryhl Dec 9 '14 at 21:15
• @KristofferRyhl My bad, thanks. Calculated too fast and wanted the discriminant to be nice square, that's why I ended up with $\frac{4}{25} + 8 = \frac{169}{25}$, so off. :-) – quapka Dec 9 '14 at 21:48

Notice that it is not just $\sqrt{51}$, it is $\frac25\sqrt{51}$. You are right in getting $8+\frac{4}{25}=\frac{204}{25}$, but the given solutition takes this farther, factoring to get $\frac{204}{25}=51\cdot\frac{4}{25}=51\cdot\left(\frac{2}{5}\right)^2$. This means that $\sqrt{\frac{204}{25}}=\sqrt{51\cdot\left(\frac{2}{5}\right)^2}=\frac{2}{5}\sqrt{51}$.
Clear denominator: $5y^2 - 2y - 10 = 0 \Rightarrow y = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot 5\cdot (-10)}}{2\cdot 5} = \dfrac{2 \pm \sqrt{204}}{10} = \dfrac{2 \pm 2\sqrt{51}}{10} = \dfrac{1 \pm \sqrt{51}}{5}$.