Hi,
or, if you prefer to "be modern", you can try
\startformula
f(x)
\alignhere = a\left(x²+\frac{b}{a}x\right)+c
\breakhere = a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c
\breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c
\breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}
\breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}
\breakhere = a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}
\numberhere[eq:foo]
\stopformula
/Mikael
On Sun, Aug 31, 2025 at 12:24 PM Mikael Sundqvist <[email protected]> wrote:
>
> Hi,
>
> On Sun, Aug 31, 2025 at 12:08 PM Fabrice Couvreur
> <[email protected]> wrote:
> >
> > Hi,
> > In the example below, the formula number is not located on the baseline of
> > the last equality. How can I make this happen ?
> > Thanks for your help.
> > Fabrice
> >
> >
> > \starttext
> > \startplaceformula
> > \startformula
> > \startalign
> > \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR
> > \NC \EQ
> > a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR
> > \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR
> > \NC \EQ
> > a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR
> > \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR
> > \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR
> > \stopalign
> > \stopformula
> > \stopplaceformula
> > \stoptext
>
> Add it to the last \NR:
>
> \starttext
> \startplaceformula
> \startformula
> \startalign
> \NC f(x) \EQ a\left(x²+\frac{b}{a}x\right)+c\NR
> \NC \EQ
> a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a²}\right)+c\NR
> \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+c\NR
> \NC \EQ
> a\left(x+\frac{b}{2a}\right)^2-\frac{b²}{4a}+\frac{4ac}{4a}\NR
> \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{b²-4ac}{4a}\NR
> \NC \EQ a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a}\NR[eq:foo]
> \stopalign
> \stopformula
> \stopplaceformula
> \stoptext
>
> /Mikael
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