The number of real solutions of the equation $x^2-9x+20=0$  is

$a=1,b=-9,c=20$

$\Delta =b^2-4ac$

$={\left(-9\right)}^2-4\left(1\right)\left(20\right)$

$=81-80$

$=1>0$

It has two real solutions

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{9\pm \sqrt{1}}{2\times 1}$             $\left[\because b^2-4ac=1\right]$

$=\frac{9\pm 1}{2}$

$=\frac{9+1}{2}or\frac{9-1}{2}$

$=5or4$                    (two real solution)