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There are two consecutive odd integers whose product is$483.$

We have to find the integers.

Let the first integer be$x$

and next consecutive odd integer be$x+2$

Therefore, the product of two consecutive odd integers is$\left(x\right)(x+2)=483.$

The equation we get is

$$\begin{gathered} \left(x\right)(x+2)=483 \\ {x}^2+2x-483=0 \\ {x}^2+23x-21x-483=0 \\ x(x+23)-21(x+23)=0 \\ (x-21)(x+23)=0 \end{gathered}$$

So,

$$\begin{gathered} x-21=0 \\ x=21 \end{gathered}$$and $$\begin{gathered} x+23=0 \\ x=-23 \end{gathered}$$

If $x=21$

So, the first integer is$21$

And the next consecutive odd integer is

role="math" localid="1644589927253" $21+2=23.$

If $x=-23$

So, the first integer is $-23$

And the next consecutive odd integer is

role="math" localid="1644590009758" $-23+2=-21.$

As the product of two consecutive odd integers is$483.$

Let's verify the integers we get by multiplying them.

So, the two consecutive odd integers are $21,23$and their product is$21脳23=483.$

And another two consecutive odd integers are $-23,-21$and their product is$(-23)脳(-21)=483.$

Thus, both pairs of consecutive odd integers are solutions.