Решить формулы корней квадратного уравнения:

1)

(4 х-5) ^2 - (2x+3) ^2=0

2)

(x^2+4x+11) ^2 = (7x^2+2x+3) ^2

1) (4x-5) ^2 - (2x+3) ^2 = (4x-5+2x+3) * (4x-5-2x-3) = (6x-2) * (2x-8) = 0

6x-2=0

x=1/3

2x-8=0

x=4

Отв: 1/3; 4

2) (x^2+4x+11+7x^2+2x+3) (x^2+4x+11-7x^2-2x-3) = (8x^2+6x+14) (-6x^2+2x+8) = 0

8x^2+6x+14=4x^2+3x+7=0

D=9-112<0 решения нет

3) - 6x^2+2x+8=3x^2-x-4=0

D=1+48=49

x1 = (1+7) / 6=4/3

x2 = (1-7) / 6=-1

Отв: 4/3; -1

1)

(4x-5) ^2 - (2x+3) ^2 = (4x-5+2x+3) * (4x-5-2x-3) = (6x-2) * (2x-8) = 0

6x-2=0

x=1/3

2x-8=0

x=4

2) (x^2+4x+11+7x^2+2x+3) (x^2+4x+11-7x^2-2x-3) = (8x^2+6x+14) (-6x^2+2x+8) = 0

8x^2+6x+14=4x^2+3x+7=0

D=9-112<0 решения нет

-6x^2+2x+8=3x^2-x-4=0

D=1+48=49

x1 = (1+7) / 6=4/3

x2 = (1-7) / 6=-1