(5x-6)^1/2=x

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Before solving the equation, we'll impose conditions of existence of the square root.

5x-6 >= 0

We'll subtract 6 both sides:

5x >= 6

We'll divide by 5:

x >=6/5

The interval of admissible solutions for the given equation is:

[6/5 , +infinite)

Now, we'll solve the equation:

sqrt (5x-6) = x

We'll square raise both sides:

5x - 6 = x^2

We'll move all terms to one side and we'll use the symmetric property:

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25+24)]/2

x2 = [5-sqrt(25+24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

Since both values belong to the interval of admissible values, they are accepted as solutions of the given equation.

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