find the base of an isoceles triangle whose area is 48 square cm and of one of the equal sides is 10

Let the 3rd side = x

S = (x+10+10)/2 = x/2 + 10

48 = [(x/2 + 10) (x/2 + 10-10) (x/2 + 10-10) (x/2 + 10 - x)]

48 = (x/2 + 10)(x/2)(x/2)(-x/2 + 10)

48 = (100 - x²/4)(x²/4)

(100 - x²/4)(x²/4) = 48²

400x²/16 - x⁴/16 = 2304

400x² - x⁴ = 36864

x⁴-400x² +36864 = 0

(x²-256)(x²-144) = 0

x² = 256 or x² = 144

x = 16 or x = 12