Solution:-

Let us assume the two numbers be A and B, B being the bigger number.

As per the condition given in the question,

$A + B = 25$ … [equation (i)]

\[2{{B}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\] … [equation (ii)]

Now, consider the equation (i),$A + B = 25$

$B = 25 – A$

Then, substitute the value of B is equation (ii),

\[2{{\left( 25-A \right)}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\]

We know that, \[{{\left( a-b \right)}^{2}}~=\text{ }{{a}^{2}}~-2ab\text{ }+\text{ }{{b}^{2}}\]

\[2({{25}^{2}}-\left( 2\text{ }\times \text{ }25\text{ }\times \text{ }A \right)\text{ }+\text{ }{{A}^{2}})\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]

\[1250\text{ }+\text{ }2{{A}^{2}}-100A\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]

\[{{A}^{2}}~+\text{ }100A-1221\text{ }=\text{ }0\]

\[{{A}^{2}}~-11A\text{ }+\text{ }111A-1221\text{ }=\text{ }0\]

Take out common in each terms,

$A(A – 11) + 111(A – 11) = 0$

$(A – 11) (A + 111) = 0$

Equate both to zero,

$A – 11 = 0$

$A + 111 = 0$

$A = 11$

$A = -111$

Therefore, $A = 11$ … [because A can’t be negative]

So, $B = 25 – A$

$= 25 – 11$

$= 14$