Решите уравнение х+1)(х-3)+20(х-1)(х+1)=3(3х-2)²+13

(a-b)²=a²-2ab+b²
a²-b²=(a-b)(a+b)

(7х+1)(х-3)+20(х-1)(х+1)=3(3х-2)²+13

(7х+1)(х-3)=7x²+x-21x-3=7x²-20x-3
(х-1)(х+1)=x²-1²=x²-1
(3х-2)²=3²x²-2*3x*2+2²=9x²-12x+4

(7х+1)(х-3)+20(х-1)(х+1)=3(3х-2)²+13
(7x²-20x-3)+20(x²-1)=3(9x²-12x+4)+13
7x²-20x-3+20x²-20=27x²-36x+12+13
7x²+20x²-27x²-20x+36x=12+13+3+20
16x=48
x=48÷16
x=3