**√(3x-2)=1+√(2x-3) solve two radicals equation?**

**Answer:**

Already on opposite sides so just square both sides

(3x - 2) = 1 + 2√(2x - 3) + (2x - 3)

Solve for the radical term

3x - 2x - 2 - 1 + 3 = 2√(2x - 3)

Combine like terms

x = 2√(2x - 3)

Square both sides again

x² = 4(2x - 3) = 8x - 12

Standard form

x² - 8x + 12 = 0

Factor

(x - 2)(x - 6) = 0

Solve

x = 2 or 6

Test each solution and both work, so the solution remains unchanged