Verified answer

Объяснение:

[tex]\displaystyle\\\frac{5x-7}{x-3} =\frac{4x-3}{4}[/tex]

ОДЗ: х-3≠0        х≠3.

[tex]\displaystyle\\4*(5x-7)=(4x-3)*(x-3)\\\\20x-28=4x^2-12x-3x+9\\\\4x^2-35x+37=0\\\\D=(-35)^2-4*4*37=1225-529=633.\\\\x_1=\frac{35-\sqrt{633} }{2*4} =\frac{35-\sqrt{633} }{8} \approx1,23.\\\\x_2=\frac{35+\sqrt{633} }{8} \approx7,52.[/tex]