Explanation:
the solution approach is described just above b) :
EA² = EC × ED
the exact same thing is to be applied in b)
just the letters of the alphabet for the names of the points are different.
so, please map them. this is just common sense. no special math "trickery" is needed.
V = A
W = E
Y = C
Z = D
x = WY = EC
WZ = ED = 15 + x
so, do you see it now ?
10² = x × (x + 15)
100 = x² + 15x
x² + 15x - 100 = 0
remember, for a quadratic equation
ax² + bx + c = 0
the general solution is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 15
c = -100
x = (-15 ± sqrt(15² - 4×1×-100))/(2×1) =
= (-15 ± sqrt(225 + 400))/2 =
= (-15 ± sqrt(625))/2 = (-15 ± 25)/2
x1 = (-15 + 25)/2 = 10/2 = 5
x2 = (-15 - 25)/2 = -40/2 = -20
a negative solution does not make any sense for actual lengths, so, x = 5 is our solution.