find the equation of the tangent line?

Differentiating it, you get:

3x^2 + 3/2

Sub in x=1, the gradient at point (1,2) is:

3 + 3/2 = 9/2

Using Y = mX + c, sub in Y=2, m=9/2, X=1:

2 = 9/2 * 1 + c

c = -5/2

Hence the equation of the tangent line is:

y = 9/2 x - 5/2

OR

2y = 9x - 5

The point (1, 2) is invalid.

I don't know if you meant y = x^3 + (3/2)x or y = x^3 + 3/(2x), but either way, if you substitute x = 1 and y = 2, the equation is false.

Please watch your parentheses.

You must mean:

... y = (x^3 + 3) / (2x)

or dy/dx = x - 3/(2x^2) [ @ x = 1 ] = -.5 ← slope

... y = mx + b

or y = -.5x + b

or 2 = -.5(1) + b

or b = 2.5

... y = -.5x + 2.5 ← final answer in slope-intercept form

y=x^3+3/2x

y'=3x^2+3/2

3x^2=-3/2

x^2=-1/2

x=-sqrt(1/2)

y-2=-sqrt(1/2)*(x-1)

y=2-sqrt(1/2)*(x-1)