Prove that (5-2√3)2 is irrational number?

We will prove it in a  same way as we prove that $\sqrt{3}$ is an irrational number..

Let us assume that $({5-2\sqrt{3}})^2$ is a rational number. 

Then          $( {5-2\sqrt{3}})^2={p\over q}$    where p and q are co-prime.

$=>$  $25-2×5×2{\sqrt 3}+(2{\sqrt 3})^2={p\over q}$    [by using $(a-b)^2=a^2-2ab+b^2$

                             $25-20{\sqrt 3} +4×3={p\over q}$

                               $25-20{\sqrt 3}+12={p\over q}$

                                         $37-20{\sqrt 3}={p\over q}$

                                        $37+{p\over q}=20{\sqrt3 }$

                                       ${37\over 20} +{p\over 20q}={\sqrt 3}$

      Clearly L.H.S. is a sum of two rational number and therefore L.H.S  is rational.

So ${\sqrt 3}$ is a rational number.

But  we know that ${\sqrt3}$ is an irrational number.So our assumption is wrong.

Hence $({5-2\sqrt{3}})^2$is an irrational number.

prove (5-2 root3)^2 ₌ 25-20root 3+12=p/q consider it as rational first

but solving it we get that root 3 could be represented as rational but it is irrational number thus our assumption that (5-2 root 3)²was wrong and hence it is irrational