How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.

Square both sides, multiply both sides by $5$, then move all terms to the left side. Note that your original equation is really two equations, but after squaring they are the same:
\begin{align}x-\tfrac45=\pm\tfrac{\sqrt{31}}{5}
&\implies \left(x-\tfrac45\right)^2=\left(\pm\tfrac{\sqrt{31}}{5}\right)^2 \\
&\implies x^2-\tfrac85 x + \tfrac{16}{25}=\tfrac{31}{25} \\
&\implies 5x^2-8 x + \tfrac{16}{5}=\tfrac{31}{5} \\
&\implies 5x^2-8 x - \tfrac{15}{5} = 0 \\
&\implies 5x^2-8 x - 3 = 0.
\end{align}
So $a=-8$ and $b=-3$.
