\[ (x+4)(x+3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12 \]
\[ x^2 + 7x + 12 = 2 \]
\[ x^2 + 7x + 10 = 0 \]
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 1 \cdot 10 = 49 - 40 = 9 \]
\[ x_1 = \frac{-7 + \sqrt{9}}{2 \cdot 1} = \frac{-7 + 3}{2} = \frac{-4}{2} = -2 \]
\[ x_2 = \frac{-7 - \sqrt{9}}{2 \cdot 1} = \frac{-7 - 3}{2} = \frac{-10}{2} = -5 \]
Ответ: -2