## Algebra 1: Common Core (15th Edition)

$123$
$(3c^{2}-3d)^{2}-21$ Substitute $c=3$ and $d=5$ into the equation above and solve, using order of operations: $=(3(3^{2})-3(5))^{2}-21$ $=(3(9)-15)^{2}-21$ $=(27-15)^{2}-21$ $=(12)^{2}-21$ $=144-21$ $=123$