Rozwiąż równanie :  a) x² + 5x - 6 = 0  b) 3 (x² - 1) = 10x - 6 Rozwiąż nierówność :  a) (x + 1)² > 1 

a) x² + 5x - 6 = 0 Δ = 5² - 4×1×(-6) = 25 + 24 = 49 √Δ = 7 x1 = (-5-7)/2 = -12/2 = -6 x2 = (-5+7)/2 = 2/2 = 1 b) 3(x² - 1) = 10x - 6 3x² - 3 - 10x + 6 = 0 3x² - 10x + 3 = 0 Δ = (-10)² - 4×3×3 = 100 - 36 = 64 √Δ = 8 x1 = (10 - 8)/2×3 = 2/6 = 1/3 x2 = (10 + 8)/2×3 = 18/6 = 3 a) (x + 1)² > 1 x² + 2x + 1 - 1 > 0 x² + 2x > 0 x (x + 2) > 0 x = 0  ∨ x = -2 x ∈ ( -∞; -2) U (0; +∞)