Type 7:  $2x^{2}-(4+\sqrt{13})x+2\sqrt{13}$

Basic method : $2x^{2}-4x-\sqrt{13}x+2\sqrt{13}$

$2x(x-2)-\sqrt{13}(x-2)$

$(2x-\sqrt{13})(x-2)$

$2x-\sqrt{13}=0$            $x-2=0$

$x=\frac{\sqrt{13}}{2}$                          $x=2$

Shortcut method:

Step 1: Consider the middle term  $(4+\sqrt{13})$

Step 2: Split the term into 4 and $\sqrt{13}$

Step 3: Divide  4 and $\sqrt{13}$  by the first term 2.

$\frac{4}{2},\frac{\sqrt{13}}{2}=2,\frac{\sqrt{13}}{2}$

The roots of the given equation are  $2,\frac{\sqrt{13}}{2}$

$10y^{2}-(18+5\sqrt{13})y+9\sqrt{13}$

Step 1: Consider the middle term  $(18+5\sqrt{13})$

Step 2: Split the term into 18 and $5\sqrt{13}$

Step 3: Divide  18 and $5\sqrt{13}$  by the first term 10

$\frac{18}{10},\frac{5\sqrt{13}}{10}=1.8,\frac{\sqrt{13}}{2}$

The roots of the given equation are  $1.8,\frac{\sqrt{13}}{2}$

The final answer is $x\geq&space;y$