Tentukan akar persamaan kuadrat a. X2+3x+-10 b.3x2+5x+1

a.
x² + 3x - 10 = 0
(x + 5) (x - 2) = 0
x = -5
x = 2

b.
3x² + 5x + 1 = 0

x1,2 = [-b +/- akar(b² - 4ac)]/2a
x1,2 = [-5 +/- akar(5² - 4(3)(1)]/2(3)
x1,2 = [-5 +/- akar13]/6

x1 = (-5 + akar13)/6
x2 = (-5 - akar13)/6