1)

[tex](4x - 1)(x - 4) - (3x - 3)(2x + 3) = 10 - 13x \\ 4 {x}^{2} - 16x - x + 4 - 6 {x}^{2} - 9x + 6x + 9 - 10 + 13x = 0 \\ - 2 {x}^{2} - 7x + 3 = 0 \\ 2 {x}^{2} + 7x - 3 = 0 \\ a = 2 \\ b =7 \\ c = - 3 \\ D = {b}^{2} - 4ac = {7}^{2} - 4 \times 2 \times ( - 3) = \\ = 49 + 24 = 73 \\ x_{1} = \frac{ - 7 + \sqrt{73} }{2 \times 2} = \frac{ - 7 + \sqrt{73} }{4} \\ x_{2} = \frac{ - 7 - \sqrt{73} }{2 \times 2} = \frac{ - 7 - \sqrt{73} }{4} [/tex]

2)

[tex] \frac{(3x + 11) {}^{2} }{5} + \frac{(2x + 5)(x + 4)}{2} = 1 + \frac{(x + 7) {}^{2} }{5} \: \: | \times 10 \\ 2(3x + 11) {}^{2} + 5(2x + 5)(x + 4) = 10 + 2(x + 7) {}^{2} \\ 2(9x {}^{2} + 66x +1 21) + 5(2 {x}^{2} + 8x + 5x + 20) - 10 - 2( {x}^{2} + 14x + 49) = 0 \\ 18 {x}^{2} + 132x + 242 + 10 {x}^{2} + 65x + 100 - 10 - 2 {x}^{2} - 28x - 98 = 0 \\ 26 {x}^{2} + 169x + 234 = 0 \\ 2 {x}^{2} + 13x + 18 = 0 \\ a = 2 \\ b = 13 \\ c = 18 \\ D = {b}^{2} - 4ac = {13}^{2} - 4 \times 2 \times 18 = \\ = 169 - 144 = 25 \\ x_{1} = \frac{ - 13 + 5}{2 \times 2} = - \frac{8}{4} = - 2 \\ x_{2} = \frac{ - 13 - 5}{2 \times 2} = - \frac{18}{4} = - 4.5[/tex]