# Solve for r value: 2 = sqrt(3r - 2) - sqrt(10-r)

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2 = sqrt(3r - 2) - sqrt(10-r)

To solve, let us square both sides:

2^2 = [ sqrt(3r-2) - sqrt(10-r)]^2

==> 4 = (3R-2) - 2SQRT(3r-2)(10-r) + (10-r)

==> 4 = (3r-2) + (10-r) - 2sqrt(-3r^2+ 32r - 20)

==> 4= 2r + 8 - 2sqrt(-3r^2 + 32r - 20)

==> 2sqrt(-3r^2 + 32r - 20) = 2r + 4

Divide by 2:

==> sqrt(-3r^2 + 32r - 20) = r + 2

Now we will square bith sides:

==> -3r^2 + 32r - 20 = (r+2)^2

==> -3r^2 + 32r - 20 = r^2 + 4r + 4

Now combine like terms:

==> -4r^2 + 28r - 24 = 0

Now divide by -4:

==> r^2 - 7r + 6 = 0

Now factor:

==> (r -6)(r-1) = 0

**==> r1= 6**

**==> r2= 1**

First, we'll impose the constraint of existence of the square roots:

3r - 2 >=0

3r > =2

r>=2/3

10 - r>=0

r=<10

The interval of admissible values for t is [2/3 ; 10].

Now, we'll solve the equation. We'll multiply both sides by the conjugate of the expression sqrt(3r - 2) - sqrt(10-r).

2 = sqrt(3r - 2) - sqrt(10-r) (1)

2[sqrt(3r - 2) + sqrt(10-r)] = 3r - 2 - 10 + r

We'll combine like terms:

2[sqrt(3r - 2) + sqrt(10-r)] = 4r - 12

We'll divide by 2:

[sqrt(3r - 2) + sqrt(10-r)] = 2r - 6 (2)

We'll add (1) + (2):

sqrt(3r - 2) - sqrt(10-r) + sqrt(3r - 2) + sqrt(10-r) = 2 + 2r - 6

We'll combine and eliminate like terms:

2sqrt(3r - 2) = 2r - 4

We'll divide by 2:

sqrt(3r - 2) = r - 2

We'll square raise both sides:

3r - 2 = (r-2)^2

3r - 2 = r^2 - 4r + 4

We'll use the symmetric property and we'll combine like terms:

r^2 - 4r + 4 - 3r + 2 = 0

r^2 - 7r + 6 = 0

**r1 = 1**

**r2 = 6**

**Since both solutions are in the interval of admissible values, they are accepted.**