Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 42

$ y=-17$ is the minimum, when $ x=-2$

Given $$ y=3x^{2}+12 x -5 $$ So, we have \begin{aligned} y&=3x^{2}+12 x -5 \\ &=3\left(x^{2}+4 x\right)-5\\ & =3\left(x^{2}+4 x+4-4\right)-5\\ & =3\left(x^{2}+4 x+4\right)-5-12\\ & =3\left(x+2\right)^{2}-17\\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $ y=-17$ is the minimum, when $ x=-2$.