Solution 2.1:2c

We obtain the answer by using the squaring rule, \displaystyle (a+b)^2=a^2+2ab+b^2, on the quadratic term and expanding the other bracketed terms:

\displaystyle \qquad (3x+4)^2-(3x-2)(3x-8)

\displaystyle \qquad \qquad \begin{align} &=\big( (3x)^2+2\cdot 3x \cdot 4 +4^2 \big) - (3x\cdot 3x-3x\cdot 8 - 2\cdot 3x+ 2\cdot 8)\\ &= (9x^2+24x+16)-(9x^2-24x-6x+16)\\ &=(9x^2+24x+16)-(9x^2-30x+16)\\ &=(9x^2+24x+16)-9x^2+30x-16\\ &=9x^2-9x^2+24x+30x+16-16\\ &=0+54x+0\\ &= 54x \end{align}