We'll use another manner to solve the problem. We'll factorize by x to the left side and by 3 to the right side:

x^2 -5x = 3x - 15

x(x-5) = 3(x-5)

We'll subtract 3(x-5) both sides:

x(x-5) - 3(x-5) = 0

We'll factorize by (x-5):

(x-3)(x-5) = 0

We'll set each factor as zero:

x - 3 = 0

**x1 = 3**

x - 5 = 0

**x2 = 5**

**The solutions of the equation are: {3 ; 5}.**

**Another manner would have been to apply the quadratic formula for the expression x^2 - 2x + 15 = 0**

x^2-5x=3x-15. To solve for x.

Subtract 3x-15 :

x^2-5x -3x+15 = 0

x(x-5) -3(x-5) = 0

(x-5)(x-3) = 0

x-5 = 0 or x-3 = 0

x= 5 or x= 3.

The equation x^2 -5x = 3x - 15 has to be solved.

x^2 -5x = 3x - 15

x^2 - 5x - 3x + 15 = 0

x^2 - 8x + 15 = 0

This is a quadratic equation. The roots of ax^2 + bx + c = 0 are given by the formula `(-b+-sqrt(b^2 - 4ac))/(2a)`

Here, a = 1, b = -8 and c = 15

The roots are `(8+-sqrt(64 - 60))/2`

= `(8+-sqrt 4)/2`

= `(8+- 2)/2`

= `4 +- 1`

= 5 and 3

The roots of the equation are x = 3 and x = 5

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