Firstly, we need to simplify both the equations and bring all terms to one side.

√x²-9x+18 = √x² 2x-15

Squaring both sides of the equation, we get:

x² - 9x + 18 = x² (2x - 15)²

Expanding and simplifying the equation, we get:

4x³ - 39x² + 108x - 81 = 0

Similarly, for the second equation:

√x²-4x+3 = x²-4x+3

Squaring both sides of the equation, we get:

x² - 4x + 3 = x⁴ - 8x³ + 25x² - 24x + 9

Simplifying the equation, we get:

x⁴ - 8x³ + 26x² - 20x + 6 = 0



The equations are now in the form of a polynomial equation. We can use factorization and synthetic division to find the roots of the equation.

For the first equation, we can factorize the equation as:

(x-3)(4x² - 21x + 27) = 0

Using synthetic division, we can solve the quadratic equation and get the roots as:

x₁ = 3, x₂ = 5/4 + 3/4i, x₃ = 5/4 - 3/4i

For the second equation, we can factorize the equation as:

(x-1)(x-2)(x² - 6x + 3) = 0

Using synthetic division, we can solve the quadratic equation and get the roots as:

x₁ = 1, x₂ = 2, x₃ = 3 + √6, x₄ = 3 - √6



Finally, we need to verify the roots by substituting them back into the original equations. If the equation is satisfied, then the root is valid.

For the first equation, substituting x = 3, we get:

√(3)² - 9(3) + 18 = √