What is the easiest way to solve this kind of quadratic equation?

Write in the form (4h)*x^2 + (2h - 2)*x + (3 + 8k) = 0

Now remember that the if the roots of px^2 + qx + r = 0 are a, b then
a + b = -q/p and ab = r/p

In this case that gives
1/h + k = -(2h - 2)/4h
k/h = (3 + 8k)/4h

Can you now solve that pair of simultaneous equations?

EDIT. There is a fault in the answer below. If, in that method, you equate constants
3 + 8k = k/h
but using h = 1/4, k = -5/2, LHS = -17 and RHS = -10.

That method needs to recognise that the equations need not have the same coefficients but must have the same ratio of coefficients. One equation could be the other with all terms multiplied by a constant.

P.S. The actual solution is h = 2, k = -3/4 which makes the original equation
8x^2 + 2x - 3 = 0 ----> (2x - 1)(4x + 3) = 0 ----> x = 1/2 or -3/4
These are, of course 1/h and k.

4hx² + 2hx + 3= 2x - 8k
4hx^2 + 2hx -2x +3 +8k = 0
4hx^2 + (2h-2)x + (3+8k) = 0  (1)

(x - 1/h)( x -k)  = 0
x^2 - kx - 1/h x + k/h = 0  
x^2 - (k +1/h) x + k/h = 0    (2)

(1) = (2) so equating coefficients of x^2
  4h = 1 , so h = 1/4
equating coefficients of x
2h - 2 = - ( k + 1/h)
but h = 1/4
1/2 -2 = - ( k + 4)
     -3/2 = -k -4
k = -4 + 3/2
    = -5/2

So k = -5/2 & h = 1/4