If the quadratic equation $p{x}^2-2\sqrt{5}\mathit{px}+15=0$ has two equal roots, then find the value of p.

Given, $p{x}^2-2\sqrt{5}\mathit{px}+15=0$.
where, $a=p,$ $b=-2\sqrt{5}\mathit{p }$and c$=15$
For equal roots, Discriminants equals to zero.

So,
$D=b^2-4\mathit{ac}=0$
$\Rightarrow (-2\sqrt{5}p{)}^2-4\times p\times 15=0$
$\Rightarrow 20p^2-60p=0$
$\Rightarrow 20p\left(p-3\right)=0$
$\Rightarrow p=0, 3$

But p ≠ 0 because if p = 0, then the given equation is not a quadratic equation.
Hence, p = 3.