Question

C) x2-20x+99, bhaskara

Answer 1

x² - 20x + 99 = 0

Δ = b² - 4 * a * c

Δ = (-20)² - 4 * 1 * 99

Δ = 400 - 396

Δ = 4

x = - b +- √Δ / 2 * a

x = -(-20) +- √4 / 2 * 1

x = 20 +- 2 / 2

x' = 20 + 2 / 2  ⇒ x' = 22 / 2 = 11

x'' = 20 - 2 / 2  ⇒ x'' = 18 / 2 = 9