Thermochemistry Specific Heat Problem

A mass of 130. g of a metal is heated to 135 °C and placed in 250. mL of water at 21.3°C. If the final temperature of the water is 45.8 °C, calculate the specific heat of the metal in J/(°C g).

Please help me with this thermochemistry problem. The correct answer is 2.21 J/g-K

I have gotten this far
specific heat = ?J / ((130g x (45.8 C -21.3 C))

heat lost by metal = heat gained by water
130g metal x C x deltaTmetal = 250g water x 4.184J/gºC x deltaTwater
deltaT metal = 135 - 45.8 = 89.2ºC
deltaT water = 45.8 - 21.3 = 24.5ºC
130g x C x 89.2ºC = 250g x 4.184J/gºC x 24.5ºC
C metal = 2.21J/gºC

http://www.chemteam.info/Thermochem/Dete…

(130) (135 - 45.8) (x) = (250) (45.8 - 21.3) (4.184)