решить уравнения a) (3x+4)(4x-3)-5=(2x)+5)(6x-7) b) 4(30-2y)-y(30-2y)=0

а) (3x+4)(4x-3)-5=(2x+5)(6x-7)

12x^2 -9x + 16x - 12 - 5 = 12x^2 - 14x + 30x - 35

12x^2 - 9x + 16x - 12 - 5 - 12x^2 + 14x -30x + 35 = 0

-9x + 18 = 0 | *(-1)

9x - 18 = 0

9x = 18

x = 18:9

x  = 2

Ответ: x = 2

б) 4(30 - 2y) - y(30 - 2y) = 0

120 - 8y - 30y + 2y^2 = 0

2y^2 - 38y + 120 = 0 | :2

y^2 - 19y + 60 = 0

D = b^2 - 4ac = 361 - 240 = 121

√D = 11

y1 = 19-11 \ 2 = 8\2 = 4

y2 = 19+11 \ 2 = 30\2 = 15

Ответ : y1 = 4; y2 = 15