(i) LHS = (x – 2)^{2} + 1 = x^{2} – 4x + 4 + 1 = x^{2} – 4x + 5

Therefore, (x – 2)^{2} + 1 = 2x – 3 can be rewritten as

x^{2} – 4x + 5 = 2x – 3

i.e., x^{2} – 6x + 8 = 0

It is of the form ax^{2} + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(ii) Since x(x + 1) + 8 = x^{2} + x + 8 and (x + 2)(x – 2) = x^{2} – 4

Therefore, x^{2} + x + 8 = x^{2} – 4

i.e., x + 12 = 0

It is not of the form ax^{2} + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(iii) Here, LHS = x (2x + 3) = 2x^{2} + 3x

So, x (2x + 3) = x^{2} + 1 can be rewritten as

2x^{2} + 3x = x^{2} + 1

Therefore, we get x^{2} + 3x – 1 = 0

It is of the form ax^{2} + bx + c = 0.

So, the given equation is a quadratic equation.

(iv) Here, LHS = (x + 2)^{3 }= x^{3} + 6x^{2} + 12x + 8

Therefore, (x + 2)^{3} = x^{3} – 4 can be rewritten as

x^{3} + 6x^{2} + 12x + 8 = x^{3} – 4

i.e., 6x^{2} + 12x + 12 = 0 or, x^{2} + 2x + 2 = 0

It is of the form ax^{2} + bx + c = 0.

So, the given equation is a quadratic equation.

(i) `(x-2)^2+1 = 2x-3`

`=>x^2+4-4x+1 = 2x-3`

`=>x^2-6x+8 = 0`

As coefficient of `x^2` is not `0`, so it is a quadratic equation.

(ii)`x(x+1)+8 = (x+2)(x-2)`

`=>x^2+x+8 = x^2-4`

`=>x+12 = 0`

As coefficient of `x^2` is `0`, so it is not a quadratic equation.

(iii) `x(2x+3) = x^2+1`

`=>2x^2+3x = x^2+1`

`=>x^2+3x-1 = 0`

As coefficient of `x^2` is not `0`, so it is a quadratic equation.

(iv) `(x+2)^3 = x^3-4`

`=>x^3+2^3+3(2)(x)(x+2) = x^3-4`

`=>x^3+8+6x^2+12x = x^3-4`

`=>6x^2+12x+12 = 0`

`=>x^2+2x+2 = 0`

As coefficient of `x^2` is not `0`, so it is a quadratic equation.