Question

(3x-2) (3x+2)-(x-3) (x+2)=32

Answer 1

$\displaystyle (3x-2)(3x+2)-(x-3)(x+2)=32\\9x^2-4-(x^2+2x-3x-6)=32\\9x^2-4-(x^2-x-6)=32\\9x^2-4-x^2+x+6=32\\8x^2+2+x=32\\8x^2+2+x-32=0\\8x^2-30+x=0\\8x^2+x-30=0\\8x^2+16x-15x-30=0\\8x(x+2)-15(x+2)=0\\(x+2)(8x-15)=0\\x+2=0,8x-15=0\\\\x_1=-2,x_2=\frac{15}{8}=1,875$