You need to equate side limits as x comes near to constant 1 such that:

`lim_(x-gt1,xlt1) (x-a)/3 + 1 = lim_(x-gt1,xgt1) sqrt(2x^2-3ax+2)`

You need to substitute 1 for x both sides:

`(1-a)/3 + 1 = sqrt(2 - 3a + 2)`

`(1-a+3)/3 = sqrt(4-3a) =gt (4-a)/3 = sqrt(4-3a)`

You need to remove the square root, hence you need to raise to square both sides such that:

`((4-a)^2)/9 = 4 - 3a`

You need to expand binomial and you need to multiply by 9 both sides such that:

`16 - 8a + a^2 = 36 - 27a`

Moving all terms to the left side yields:

`a^2 + 19a - 20= 0`

You may write the constant term -20 as `-19-1` such that:

`a^2 + 19a - 19 - 1 = 0`

You need to group the terms such that:

`(a^2 - 1) + 19(a-1) = 0`

You should substitute the difference of squares such that:

`(a-1)(a+1) + 19(a-1) = 0 `

Factoring out `a-1` yields:

`(a-1)(a+1+19) = 0`

`a-1 = 0 =gt a = 1`

`a+20 = 0 =gt a = -20`

**Hence, evaluating the values of a under given conditions yields `a_1 = 1` and `a_2 = -20` .**