Укажите область значений функции y=0,5x²-1

1) 6x^2+5x-11=0
D=289
x1=-5+17/12 = 1
x2=-5-17/12 = 
ответ:1, -1 5/6

2) 8x^2+x-7=0
D=289
x1=-1+17/26 = 
x2=-1-17/26 = 
ответ: 8/13, -9/13

3) 4x^2-x-14=0
D=225
x1=1+15/8 = 2
x2=1-15/8 = 
ответ: 2, -1 3/4

4) 9x^2-14x+5=0
D=16
x1=14+4/18 = 1
x2=14-4/18 = 
ответ: 1; 5/9

5) 10x^2-14x = 2x^2-5
8x^2-14x+5=0
D=36
x1=14+6/16 = 
x2=14-6/16 = 
ответ: 1 1/4, 1/2.

6) x^2+8x+16 = 4x^2+5
3x^2-8x-11 = 0
D=196
x1=8+14/6 = 22/6 = 11/3
x2=8-14/6 = -1
ответ: 11/3; -1

7) x^2-10x+25=3x^2-x+14
2x^2+9x-11=0
D=169
x1= -9+13/6=
x2= -9-13/6=
ответ: -2/3, -11/3

8) 36x^2-9x=3x-1
36x^2-12x+1=0
D=0
x=
ответ: 1/6.

B^2 - 9y^2 : 4b^2 +12by+9y^21) 5(x+4y)/ 15(x+4y) = 1/3
2) 8a^2+40ab : ab+5b^2 = 8a (a + 5b) / b(a+5b) = 8a/b
3) 8ab : 4a^2 b^2+8ab^3 = 8ab / 4ab^2 (a +2b) = 2/b (a+2b)
4) x^2 -9 : 4x-12 = (x-3)(x+3) / 4(x-3) = x+3/4
5) 3a+15b : a^2 - 25b^2 = 3(a+5b) / (a-5b)(a+5b) = 3/a+5b
6)  y^2+4y+4 : y^2 +2y = (y+2)^2 / y(y+2) = y+2/y
7) b^2 +10b+25 : b^2 -25 = (b+5)^2 / (b-5)(b+5) = b+5/b-5

4b^2 - 9y^2 : 4b^2 +12by+9y^2 = (2b-3y)(2b+3y)/ (2b+3y)^2 = 2b-3y/2b+3y = 2*(-1/4) - 3* (5/6) / 2*(-1/4) - 3*(-5/6) = -0.5 + 2.5 / -0.5 +2.5 = 2/2 = 1