First, in order for this to makes sense the different units must measure the same thing! That is the case here- both "cubic feet" and "gallons" measure volume, both "seconds" and "minutes" measure time. "cubic feet per second" and "gallons per minute" measure how fast volume is changing- perhaps how fast a liquid is moving through a pipe.
h, in the first equation is in feet and Q is in cubic feet per second. But Q is squared so [itex]Q^2[/itex] has units of "cubic feet squared" (or feet to the sixth power) over "seconds squared". That tells us that the "10" must have units of feet and the "4.43" must have units of "seconds squared over feet to the fifth power": [itex]\frac{s^2}{ft^5}\times\frac{ft^6}{s^2}= ft[/itex] which can then be added to the "10 ft" to get "h ft".
In the second equation, h is still in feet but now Q is in gallons per minute so the "[itex]2.2*10^{-5}[/itex]" must have units of [itex]\frac{min^2 ft}{gal^2}[/tex] so that [itex]\frac{min^2 ft}{gal^2}\times\frac{gal^2}{min^2}= ft[/itex] again. There are about 7.48 cubic feet per gallon (I had to look that up) and 60 seconds per minute (that I did not).<br />
That gives [itex]\frac{1}{7.48} \frac{gal}{ft^3}[/itex] and [itex]\frac{1}{60} \frac{min}{sec}[/itex] so [itex]\frac{1}{7.48^2} \frac{gal^2}{ft^6}[/itex] and [itex]\frac{1}{3600}\frac{min^2}{sec^2}[/itex] So our conversion, from [itex]\frac{s^2}{ft^5}[/itex] to [itex]\frac{min^2 ft}{gal^2}[/itex] can be done by<br />
[tex]\left(4.43\frac{sec^2}{ft^5}\right)\times\left(\frac{1}{60}\frac{min}{sec}\right)^2\left(\frac{1}{7.48}\frac{gal}{ft^3}\right)^2= \left(4.43\right)\left(\frac{1}{3600}\right)s\left(\frac{1}{55.95}\right)= 2.18\times 10^{-5}[/tex]<br />
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That's not exactly the same as the given "[itex]2.2\times 10^{-5}[/itex]", perhaps because of round-off error, but that's the idea.[/itex]