Phosphoric Acid Problem
Let's list what's going on here. Orthophosphoric acid dissociates stepwise:

\begin{align}
\ce{H3PO4 + H2O &amp; &lt;=&gt;[$K_\mathrm{a1}$] H3O+ + H2PO4–} \tag{R1}\\
\ce{H2PO4- + H2O &amp; &lt;=&gt;[$K_\mathrm{a2}$] H3O+ + HPO4^2–} \tag{R2}\\
\ce{HPO4^2- + H2O &amp; &lt;=&gt;[$K_\mathrm{a3}$] H3O+ + PO4^3–} \tag{R3}
\end{align}

Mass balances for protons and phosporous, respectively:

\begin{align}
C_\ce{H} &amp; = [\ce{H3O+}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + [\ce{H3PO4}] = [\ce{H3O+}] + \sum_{n=1}^3 [\ce{H_$n$PO4^$(3-n)-$}] \tag{1}\\
C_\ce{P} &amp; = [\ce{PO4^3-}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + \ce{H3PO4} \tag{2}
\end{align} 

Now we compare the equilibrium constants:

\begin{align}
K_\mathrm{a1} &amp; = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.1 \times 10^{−3} \tag{3}\\
K_\mathrm{a2} &amp; = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.3 \times 10^{−8} \tag{4}\\
K_\mathrm{a3} &amp; = \frac{[\ce{H3O+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 4.2 \times 10^{−13} \tag{5}
\end{align}

Notice that $K_\mathrm{a1} &gt;&gt; K_\mathrm{a2}$ (more that $10^5$), therefore by the first dissociation step phosphoric acid behaves as an acid of medium strength, producing equal amounts of $\ce{H3O+}$ and $\ce{H2PO4-}$ (considering charge balance): 

$$[\ce{H3O+}] \approx [\ce{H2PO4-}] \tag{6}$$

which directly leads us to the answer. Using expression for $K_\mathrm{a2}$:

$$[\ce{HPO4^{2-}}] = K_\mathrm{a2} \times \frac{[\ce{H2PO4-}]}{[\ce{H3O+}]} = K_\mathrm{a2} \tag{7}$$

That's it, $[\ce{HPO4^{2-}}] = \pu{6.3 \times 10^{−8} (mol/L)}$



Now for the fun part, if you really want to solve quadratic equation, we can also find solution pH and $[\ce{PO4^{3-}}]$.

Based on the values of equilibrium constants, it is safe to assume that pretty much only first dissociation step defines solution's pH. We can rewrite equation for $K_\mathrm{a1}$ as follows:

$$K_\mathrm{a1} = \frac{[\ce{H3O+}]^2}{C_0 - [\ce{H3O+}]},$$

where $C_0$ is the concentration of the solution ($\pu{0.10 M}$). Using the given numbers we have

$$7.1 \times 10^{-3} = \frac{[\ce{H3O+}]^2}{0.10 - [\ce{H3O+}]}$$
$$[\ce{H3O+}]^2 + 7.1 \times 10^{-3} [\ce{H3O+}] - 7.1 \times 10^{-4} = 0 $$ 
$$[\ce{H3O+}] = \frac{-7.1 \times 10^{-3} + \sqrt{(7.1 \times 10^{-3})^2 + 4 \times 7.1 \times 10^{-4}}}{2} = \pu{2.33 \times 10^{-2} (mol/L)}$$

$$\mathrm{pH} = - \log{[\ce{H3O+}]} = 1.63$$

WolframAlpha thinks the same.

Now we know $[\ce{H2PO4-}] = \pu{2.33 \times 10^{-2} (mol/L)}$ and we can finally find $[\ce{PO4^3-}]$ from the expression for $K_\mathrm{a3}$:

$$[\ce{PO4^{3-}}] = K_\mathrm{a3} \times \frac{[\ce{HPO4^2-}]}{[\ce{H3O+}]} = 4.2 \times 10^{−13} \times \frac{6.3 \times 10^{−8}}{2.33 \times 10^{-2}} = \pu{1.1 \times 10^{-18} (mol/L)}$$

This demonstrates that in pretty concentrated solution of phosphoric acid anions $\ce{PO4^3-}$ are practically non-existing.