Multiplying

Multiply complex numbers the way you multiply two brackets: every term of one times every term of the other. For $\left(2+i\right)\left(3+4i\right)$ that gives $6+8i+3i+4{i}^{2}$.

Now the one new step: ${i}^{2} = -1$, so $4{i}^{2}$ is -4. The line becomes $6+11i-4$, and collecting gives $2+11i$.

Why replace ${i}^{2}$? It is a real number in disguise, and replacing it is what brings the answer back to the form $a+bi$. Watch the signs: a product like $3i\cdot \left(-2i\right)$ is $-6{i}^{2}$, which is +6.