Verified answer

[tex]f'=\frac{(5*2x-3)(x-3)-(5x^{2}-3x-1)*1 }{(x-3)^{2} } =\frac{10x^{2}-33x+9-5x^{2}+3x+1 }{(x-3)^{2}}[/tex] =

=[tex]\frac{5x^{2}-30x+10}{(x-3)^{2}} =\frac{5(x^{2}-6x+2)}{(x-3)^{2}}[/tex]

[tex]\frac{5(x^{2}-6x+2)}{(x-3)^{2}}=0[/tex] одз х≠3

х²-6х+2=0 ,D=28=4*7

[tex]x_1=\frac{6-2\sqrt{7} }{2} =3-\sqrt{7} \\[/tex] ≠3

[tex]x_2=\frac{6+2\sqrt{7} }{2} =3+\sqrt{7}[/tex]≠3

Ответ .[tex]3-\sqrt{7} , 3+\sqrt{7} \\[/tex]