x+√x -20=0 x^5-9x^3+20x=0

1) $x \ + \sqrt{x} -\ 20\ = 0\\ t^{2}\ =\ x\\ t^{2}\ +\ t\ -\ 20\ =\ 0\\ D\ =\ 1+\ 80\ =\ 81\ =\ \sqrt{9}\\ t_{1}\ =\ -5\\ t_{2}\ =\ 4\\ x_{1}\ =\ 25\\ x_{2}\ =\ 16$

2) $x^{5}\ -\ 9x^{3}\ +20x\ =0\\ x(x^{4}\ -\ 9x^{2}\ +20)\ =\ 0\\ x\ =\ 0\\ x^{4}\ -\ 9x^{2}\ +20\ =\ 0\\ t\ =\ x^{2}\\ t^{2}\ -\ 9t\ +20\ =\ 0\\ D\ =\ 81-\ 80\ =\ 1\\ t_{1}\ =\ 4\\ t_{2}\ =\ 5\\ x_{1}\ =\ 2\\ x_{2}\ =\ \sqrt{5}\\$