Помогите решить!

(3x+2) ^2=12x+40

(x-1) (x+1) = 2 (x-2) (x+2)

48-3 (x^2-5) ^2=0

Найдите корни уравнения, 8 класс

1) 9x²+12x+4-12x-40=0

9x²-36=0

x²-4=0

(x-2) (x+2) = 0

x1 = - 2, x2 = 2

2) x²-1=2 (x²-4)

x²-1-2x²+8=0

-x²+7=0

x²-7=0

(x-√7) (x+√7) = 0

x1 = - √7, x2 = √7

3) 48-3 (x²-5) ²=0

16 - (x²-5) ²=0

(4 - (x²-5)) * (4 + (x²-5)) = 0

(4-x²+5) (4+x²-5) = 0

(9-x²) (x²-1) = 0

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

x1=3, x2 = - 3, x3=1, x4 = - 1