1) 6x+12x²=0; 2) x²-4x-9=0; 3) 40x²+2x+3=0; 4) 6x²-11=7x²-5;

Решите задачу:

$1) \\ 6x + 12 {x}^{2} = 0 \\ 6x(1 + 2x) = 0 \\ x1 = 0 \\ x2 = - \frac{1}{2} \\ \\ 2) \\ {x}^{2} - 4x - 9 = 0 \\ d = 16 + 4 \times 9 = 16 + 36 = 52 = 4 \times13 \\ x1 = \frac{4 + 2 \sqrt{13} }{2} = 2 + \sqrt{13} \\ x2 = 2 - \sqrt{13} \\ \\ 3) \\ 40 {x}^{2} + 2x + 3 = 0 \\ d = 4 - 4 \times 40 \times 3 < 0 \\ \\n et \: korney \\ \\ 3) \\ 6 {x}^{2} - 11 = 7 {x}^{2} - 5 \\ 7 {x}^{2} - 6 {x}^{2} = - 11 + 5 \\ {x}^{2} = - 6 \\ {x}^{2} = - 6 \\ net \: korney$