Persamaan kuadrat (m-3)x^2-4x+m = 0 mempunyai dua akar-akar real. batas-batas nilai m yang memenuhi adalah...

→(m-3)x² - 4x + m = 0←
a = (m-3) , b = -4 , c = m

b² - 4ac = 0
(-4)² - 4(m-3)(m) = 0
16 - (4m-12)m = 0
16 - (4m² - 12m) = 0
-4m² + 12m + 16 = 0
4m² - 12m - 16 = 0
m² - 3m - 4 = 0
(m+1)(m-4) = 0
→m = -1 atau m = 3