Find three consecutive multiples of 5 so that the square of the third number, decreased by 5 times the second number is the same as 25 more than twice the product of the first two numbers

Let the unknown value be 'x'

Consider the three consecutive multiples of 5 as 'x', '(x + 5)' and '(x + 10)'

First number = x
Second number = (x + 5)
Third number = (x + 10)

Using the above statement, "The square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers," let us form an equation

(x + 10)² - 5(x + 5) = 25 + (2 (x) (x + 5) )
x² + 20x + 100 - 5x - 25 = 25 + 2x² + 10x
x² + 15x + 75 = 2x² + 10x + 25

Bringing everything to one side,

0 = 2x² + 10x + 25 -x² -15x -75
or
2x² + 10x + 25 -x² -15x -75 = 0
x² - 5x - 50 = 0

Factorising the above equation, 
(x - 10) (x + 5) = 0 ⇒ x = 10 or x = -5

If x = 10, then the three consecutive numbers are 10, 15, 20.

Let us check if it follows the above statement!

Square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers
20² - (5 x 15) = 25 + (2 x (10 x 15))
400 - 75 = 25 + (2 x 150)
325 = 25 + 300
325 = 325

Now let us try with x = -5. The three consecutive numbers are -5, 0, 5

Let us check if it follows the above statement.

Square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers
5² - (5 x 0) = 25 + (2 x (-5 x 0))
25 - 0 = 25 + (2 x 0)
25 = 25 + 0
25 = 25

Hence both x = 10 and x = -5 are correct

9/12 is the answer because 3/4 times 3 each is 9/12