(3x-1)(4x+2)-(5x+3)^2= 38 ПОМОГИТЕ ПОЖАЛУЙСТА!

Решите задачу:

$(3x - 1)(4x + 2) - (5 + 3)^{2} = 38 \\ 12x^{2} + 6x - 4x - 2 - 25x^{2} - 30x - 9 = 38 \\ 12x^{2} - 25x^{2} - 30x + 6x - 4x = 38 + 2 + 9 \\ - 13x ^{2} - 28x = 49 \\ - 13x ^{2} - 28x - 49 = 0 \\ 13x^{2} +28x + 49 = 0 \\ d = \sqrt{ {b}^{2} - 4ac } = \sqrt{ 784 - 4 \times 13 \times 49} = \sqrt{ 784 - 2548} = \sqrt{ - 1764 } \\ x1 = \frac{ - 1 + \sqrt{1764} }{26} \\ x2 = \frac{ - 1 - \sqrt{1764} }{26}$