Solution:-

Let us assume the two numbers be, P and Q.

As per the condition given in the question,

\[{{P}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }208\] … [equation (i)]

\[{{Q}^{2}}~=\text{ }18P\] … [equation (ii)]

Consider the equation (i), \[{{P}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }208\]

\[{{Q}^{2}}~=\text{ }208-{{P}^{2}}\]

Now, substitute the value of \[{{Q}^{2}}\]  in equation (ii) we get,

\[208-{{P}^{2}}~=\text{ }18P\]

By transposing we get,

\[{{P}^{2}}~+\text{ }18P\text{ }\text{ }208\text{ }=\text{ }0\]

\[{{P}^{2}}~+\text{ }26P-8P-208\text{ }=\text{ }0\]

Take out common in each terms,

$P(P + 26) – 8(P + 26) = 0$

$(P + 26) (P – 8) = 0$

Equate both to zero,

$P + 26 = 0$

$P – 8 = 0$

$P = – 26$

$P = 8$

Therefore, $P = 8$ … [because P can’t be negative]

So, \[{{Q}^{2}}~=\text{ }18P\]

\[{{Q}^{2}}~=\text{ }18\text{ }\times \text{ }8\]

\[{{Q}^{2}}~=\text{ }144\]

$Q = √144$

$Q = 12$