kalo ada yg tau tolong kasih tau yaa

(f o g)(x) = 4
(2x - 6)² - 4 = 4
4x² - 24x + 36 - 4 = 4
4x² - 24x + 28 = 0
x² - 6x + 7 = 0
x = (-b ± √(b²-4ac)) / 2(a)
x = (6 ± √(36-28)) / 2
x = (6 ± √8) / 2
x = (6 ± 2√2) / 2
x = 3 + √2 atau x = 3 - √2

Mapel : Matematika

Kelas : 10

Bab : Fungsi Komposisi
__________________

(f o g) (x) = f(2x - 6)

(f o g) (x)             = 4

(2x - 6)² - 4          = 4

(2x - 6)²               = 8

4x² - 24x + 36      = 8

4x² - 24x + 36 - 8 = 0

4x² - 24x + 28      = 0 ⇒ Sederhanakan dengan dibagi 4

x² - 6x + 7            = 0

Nilai a = 1
         b = -6
         c = 7

x = (-b ± √(b² - 4ac))/2a

x = (-(-6) ± √((-6)² - 4(1)(7)) / 2(1)

x = (6 ± √(36 - 28))/2

x = (6 ± √8)/2

x = (6 ± 2√2)/2


x = [tex] \frac{6+ 2 \sqrt{2} }{2} [/tex]
   = 3 + √2

atau

x = [tex] \frac{6-2 \sqrt{2} }{2} [/tex]
   = 3 - √2