We first turn the mathematical statement to equation i.e. \(a \sqrt{5}+b \sqrt{5}=\sqrt{95-30 \sqrt{10}}\)
we square both sides
\begin{aligned}
(a \sqrt{5}+b \sqrt{2})^{2} &=(\sqrt{95-30 \sqrt{10}})^{2} \\
a^{2} \cdot 5+2 a b \sqrt{10}+b^{2} \cdot 2 &=95-30 \sqrt{10} \\
5 a^{2}+2 b^{2}+2 a b \sqrt{10} &=95-30 \sqrt{10}
\end{aligned}
equating the terms,
\begin{aligned}
&5 a^{2}+2 b^{2}=95 \\
&2 a b \sqrt{10}=-30 \sqrt{10} \\
&a b \sqrt{10}=-15 \sqrt{10} \\
&a b=-15, a=\frac{-15}{b} \\
&\text { Sub (2) into (1) } \\
&\text { 5・( } \left.\frac{-15}{b}\right)^{2}+2 b^{2}=95
\end{aligned}
\begin{aligned}
&\frac{1125}{h^{2}}+2 b^{2}=95 \\
&\text { Let } b^{2}=P \\
&\frac{1250}{p}+2 p=95: \frac{1125+2 p^{2}}{p}=95 \\
&1125+2 p^{2}-95 p=0 \\
&\text { we rearrange } \\
&2 p^{2}-95 p+1125=0
\end{aligned}
by the formula method
\begin{aligned}
&P=\frac{95 \pm \sqrt{95^{2}-4 \times 2 \times 125}}{2 \times 2} \\
&P=\frac{95 \pm \sqrt{25}}{4} \\
&P=\frac{95 \pm 5}{4}=\frac{100}{4} \text { or } \frac{90}{4} \\
&P=25 \text { or } \frac{45}{2} \\
&\text { but } P=b^{2}
\end{aligned}
\(25=b^{2} \cdot b=\pm \sqrt{25}=\pm 5\) or \(\frac{45}{2}=b^{2}, b=\pm \sqrt{\frac{45}{2}}\) when \(b=5\). from (2),
\( a=\frac{-15}{5}=3\)
Hence, \(a=3 . b=-5\)