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@jhonyy9
\[\large\rm F=3x^2+1\qquad\qquad\qquad G=2x-3\]So if we stick G inside of F,\[\large\rm F(G)=3G^2+1\]\[\large\rm F(G)=3(2x-3)^2+1\]And then expand that all out.
Notice that when you square the first term, 4x^2, and then multiply it by 3 it gives you 12x^2. So the answer is probably one of the last two options.
so it is the third answer
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thanks a bunch !
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