+101 There exist constants $a,$ $h,$ and $k$ such that \[3x^2 + 12x + 4 = a(x - h)^2 + k\]for all real numbers $x.$ Enter the ordered triple $(a,h,k).$
+37159 So if we arrange the left side properly, we can get form on R and see what a h k are
3x^2 + 12x + 4 divide by 3
3 (x^2 + 4x) + 4 'complete the square'
3 ( x^2 +4x + 4) + 4
3(x+2)^2 + 12 + 4
3 (x+2)^2 + 16
