quadratic equation

Let x be the tens digit.
Then the unit digit is x+5 and the two-digit number is 10x + (x + 5) = 11x + 5

From the condition given by the question,
(11x + 5) (x + x + 5) = 243
(11x + 5) (2x + 5) = 243
22x² + 65x + 25 = 243
22x² + 65x - 218 = 0
(22x + 109) (x - 2) = 0
x = -109/22 (Rejected) or x = 2

So the tens digit is 2 and the unit digit is 2+5 = 7
So the two-digit number is 27.


Hope it helps! ^^

let the number be 10x+y, where x and y are positive integers smaller than 10

then
y-x=5 --> y=x+5 --------------- 1)

(10x+y)*(x + y) = 243 ---------------- 2)

Put 1) into 2)

(10x+x+5)*(x+x+5)=243
(11x+5)*(2x+5)=243
22x^2+65x+25=243
22x^2+65x-218=0
(22x+109)(x-2)=0
x=2 or x=-109/22(rejected, since x must be a positive integer)

when x=2, y=2+5=7
Therefore, the two-digit number is 27

let the ten's digit be x, the unit digit be y:
y-x=5____(1)
(10x+y)(x+y)=243____(2)
By(1), y=x+5____(3)
Sub (3) into (2),
(11x+5)(2x+5)=243
22x^2+ 65x- 218=0
=>x=2
=>y=7
i.e. The two-digit number is 27.