Пожалуйста помогите решить через дискриминант 1) 2x²-x(2√3+4)+4√3=0 2) (7x - дробь 21 /2)(х+1) 3) 77х²-65х+12=0 Прошу

1 ) 2 * x ^ 2 - x * ( 2 √3 + 4 ) + 4 √ 3 = 0 ; D = ( - ( 2 √3 + 4 ) ) ^ 2 - 4 * 2 * 4 √ 3 = 4 * 3 +16 √ 3 + 16 - 32√ 3 = 28 - 16√ 3 = (4 -2 √ 3) ^ 2 ;x1 = ( ( 2 √3 + 4 ) - (4 -2 √ 3) ) / 4 = ( 2 √3 + 4 - 4 +2 √ 3 ) / 4 = 4 √3 / 4 = √3 ;x2 = ( ( 2 √3 + 4 ) + (4 -2 √ 3) ) / 4 = ( 2 √3 + 4 + 4 - 2 √ 3 ) / 4 = 8 / 4 = 2 ;2) (7 * x - 21 / 2 ) * ( х + 1 ) = 0 ;7 * x ^ 2 + 7 * x - 10 . 5 * x - 10 . 5 = 0 ;7 * x ^ 2 - 3 . 5 * x - 10 . 5 = 0 ;D = b ^ 2 - 4ac = (-3.5) ^ 2 - 4·7·(-10.5) = 12.25 + 294 = 306.25;x1 = (3.5 - √306.25 ) /(2·7 ) = ( 3.5 - 17.5) / 14 = -14 / 14 = -1;x2 = (3.5 + √306.25 0 /(2·7) = (3.5 + 17.5 ) / 14 = 21 / 14 = 1.5;3) 77 * х ² - 65 * х + 12 = 0 ; D = b ^ 2 - 4ac = (-65) ^ 2 - 4·77·12 = 4225 - 3696 = 529;x1 = ( 65 - √529) /( 2·77 ) = (65 - 23)/ 154 = 42 /154 = 3 /11;x2 = (65 + √529)/( 2·77) = (65 + 23)/ 154 = 88 /154 = 4 /7.