{ y >_ x² – 4x – 12 6x – 3y <_ – 6 tentukan hasil penyelesaian

Y ≥ x² – 4x – 12
y ≤ -x² + 4x + 12

6x – 3y ≤ -6
-3y ≤ -6x – 6
y ≤ 2x + 2

2x + 2 ≤ -x² + 4x + 12

x² – 2x – 10 ≤ 0

x1,2 = (2 ± √2² – 4(1)(-10)/2

x1,2 = (2 ± √44)/2

x1,2 = 1 ± √11

x1 = 1 + √11

x2 = 1 – √11

Hp = { 1 – √11 ≤ x ≤ 1 + √11 }