Question

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

Answer 1

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

Answer 2

спасибо