Given: $x^2+(p-3) x+p=0$ $\because$ Roots are real and equal, then $b^2-4 a c=0$Here we compare the coefficients of $a, b$ and $c$ with the equation $a x^2+b x+c=0$ . $a=1,\ b=p-3\;\text{and}\;c=p$Now putting the values of $a, b$ and $c$ in given equation $(p-3)^2-4\times 1\times p=0$ $p^2+9-6 p-4 p=0$ $p^2+9-10 p=0$ $p^2-10 p+9=0$ $p^2-9 p-p+9=0$ $p(p-9)-1(p-9)=0$ $\Rightarrow\;\; (p-9)(p-1)=0$Hence, $p=9$ or 1