Логика такая:
\[ (x-8)(x+5) = x^2 + 5x - 8x - 40 = x^2 - 3x - 40 \]
\[ (3b-2)(4b-2) = 12b^2 - 6b - 8b + 4 = 12b^2 - 14b + 4 \]
\[ (6a+x)(2a-3x) = 12a^2 - 18ax + 2ax - 3x^2 = 12a^2 - 16ax - 3x^2 \]
\[ (c+1)(c^2+3c+2) = c^3 + 3c^2 + 2c + c^2 + 3c + 2 = c^3 + 4c^2 + 5c + 2 \]
\[ 2x(x-1) - 3(x-1) = (x-1)(2x-3) \]
\[ ab + ac + 4b + 4c = a(b+c) + 4(b+c) = (b+c)(a+4) \]
\[ (3m-n)(m^2-n^2) - 2mn(m-n) = (3m-n)(m-n)(m+n) - 2mn(m-n) = (m-n)((3m-n)(m+n) - 2mn) = (m-n)(3m^2 + 3mn - mn - n^2 - 2mn) = (m-n)(3m^2 - n^2) \]
\[ (y-5)(y+8) = y^2 + 8y - 5y - 40 = y^2 + 3y - 40 \]
\[ y(y+3) - 40 = y^2 + 3y - 40 \]
Левая часть равна правой части, следовательно, тождество доказано.
\[ 12x^2 - (4x-3)(3x+1) = -2 \]
\[ 12x^2 - (12x^2 + 4x - 9x - 3) = -2 \]
\[ 12x^2 - 12x^2 - 4x + 9x + 3 = -2 \]
\[ 5x + 3 = -2 \]
\[ 5x = -5 \]
\[ x = -1 \]
Ответ: \(x = -1\)