Question

1)(х+2)(х-3)-(2х-5)(х+3)=х(х-5)
2) (6x-5)к квадрату +(3х-2)(3х+2)=36

Answer 1

$1)quad (x+2)(x-3)-(2x-5)(x+3)=x(x-5)\\x^2-x-6-(2x^2+x-15)-x^2+5x=0\\-2x^2+3x+9=0\\2x^2-3x-9=0\\D=81; ,; ; x_1= frac{3-9}{4}=- frac{3}{2} ; ,; x_2= frac{3+9}{4} =3\\2)quad (6x-5)^2+(3x-2)(3x+2)=36\\36x^2-60x+25+9x^2-4-36=0\\45x^2-60x-15=0; |:15\\3x^2-4x-1=0\\D=28; ,\\x_1= frac{4-sqrt{28}}{6}= frac{2-sqrt7}{3} ; ,; ; x_2= frac{2+sqrt7}{3}$