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Compute inside-out. First evaluate g(5)=2⋅5+3=13g(5)=2\cdot 5 + 3 = 13. Then apply ff to that result:

f(13)=132−4(13)=169−52=117.f(13) = 13^2 - 4(13) = 169 - 52 = 117.

Thus choice C. This problem checks function composition, a recurring GRE Quant theme. The order matters: f(g(5))f(g(5)) differs from g(f(5))g(f(5)). For instance, f(5)=25−20=5f(5)=25 - 20 = 5 and g(f(5))=g(5)=13g(f(5))=g(5)=13, not 117. A helpful tip is to box the inner evaluation (here, 13) before substitution, reducing the likelihood of plugging 5 into both functions accidentally. You can also rewrite f(g(x))f(g(x)) symbolically to verify: f(2x+3)=(2x+3)2−4(2x+3)=4x2+12x+9−8x−12=4x2+4x−3f(2x+3)=(2x+3)^2 - 4(2x+3) = 4x^2 + 12x + 9 - 8x - 12 = 4x^2 + 4x - 3. Then f(g(5))=4(25)+4(5)−3=100+20−3=117f(g(5))=4(25) + 4(5) - 3 = 100 + 20 - 3 = 117, acting as a consistency check. When time is tight, numeric substitution is faster; symbolic composition is useful for verifying and for items that ask for expressions rather than values.