Question

(3x-1)(3x+1)-(x-4)(x-2)=0

Answer 1

$ax^2+2px+c=0\\\\D/4=p^2-a\cdot c\\\\x_{1,2}= \frac{-p\pm \sqrt{D/4}}{a}$

(3x-1)(3x+1)-(x-4)(x-2)=0
9x²-1-(x²-6x+8)=0
8x²+6x-9=0
D/4=81 ,  x1=-12/8=-3/2  ,  x2=6/8=3/4