Let x be the number of gallons of 50% antifreeze.
Amount of pure antifreeze added is 0.5x gallons.
Amount of pure antifreeze in 20% solution is 0.2×90=18 gallons.
Volume of final solution is x+90 gallons.
Amount of pure antifreeze in final solution is 0.4(x+90).
Mixture: 0.5x+18=0.4(x+90).
0.5x+18=0.4x+36, 0.1x=18, x=18/0.1=180 gallons.
180 gallons of 50% concentration have to be added.