When symbolically solving transcendental equations, one should appropriately use Solve or Reduce with adequate logical combinations of equations, inequalities, domain specifications, etc. to get desired results. There are limitations for certain types of equations involving transcendental functions, see e.g. Transcendental Roots. Here is an example which doesn't work in Mathematica 9:
Reduce[ Re[Zeta[x]] == 0 && 14 < Im[x] < 15 && Re[x] == 1/2, x]
Namely, it cannot find roots of the real part of Zeta; nonetheless, we can find symbolic zeros of the Riemann Zeta function:
Solve[Zeta[x] == 0 && 0 < Im[x] < 25 && Re[x] == 1/2, x]
{{x -> ZetaZero[1]}, {x -> ZetaZero[2]}}
Let's use Reduce in another way:
Reduce[ Zeta[ 1/2 + I t] == 0 && 0 <= t <= 250, t]
C[1] ∈ Integers && 1 <= C[1] <= 108 && t == -(1/2) I (-1 + 2 ZetaZero[C[1]])
Here, we provide plot of the real and imaginary part of Zeta on the critical line, in the range 0 < t < 50:
Plot[ Table[ h[ Zeta[ 1/2 + I t]], {h, {Re, Im}}], {t, 0, 50},
Evaluated -> True, AspectRatio -> 1/4, PlotStyle -> Thick,
Epilog -> {Red, PointSize[0.007], Point[{Im@ZetaZero[#], 0} & /@ Range[10]]},
ImageSize -> 800, PlotLegends -> Placed["Expressions", {Left, Top}]]

The function ZetaZero is very useful, since one can evaluate its numerical values up to 10^7-th zero (on the other hand, we know there are infinitely many Zeta zeros on the critical line), then one should work with e.g. FindRoot yielding numerical results.
ZetaZero /@ {1, 2, 10^7} // Im // N
{14.1347, 21.022, 4.99238*10^6}
If we are satisfied with numerical values only, we can use FindRoot to get zeros of the real part of Zeta between 14 and 15:
FindRoot[ Re[ Zeta[ 1/2 + I t]], {t, #, 14, 15}]& /@ {14, 14.5}
{{t -> 14.1347}, {t -> 14.5179}}
Zeta[1/2 + I t] /. % // Chop
{0, 0. + 0.31227 I}
Working with FindRoot, it should be useful to localize approximately starting values with ContourPlot like e.g. here.
Solve[Zeta[x] == -(1/12) && Abs[x] < 20, x]. – Artes Oct 23 '13 at 09:12FindRoot[Re[Zeta[1/2 + I t]], {t, #}] & /@ {14, 14.5}yields{{t -> 14.1347}, {t -> 14.5179}}. By the way your question is still far from being clear. – Artes Oct 23 '13 at 09:27Re@Zeta[1/2 + I t]?Re@Zeta[x]can intersects=1/2 + iyonly wheny == 0. – Artes Oct 23 '13 at 09:35FindRoot[Re[Zeta[1/2 + I t]], {t, #}] & /@ Range[14, 25]– Artes Oct 23 '13 at 09:37FindRootnonetheless you can get rid of them using appropriate functionality. This gives you a simple approachIm[ZetaZero[#] & /@ Range[3] // N]. – Artes Oct 23 '13 at 09:56