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

Решите задачу:

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