Solving a simple equation, avoiding a quadratic Sorry for the trivial question. 
I was wondering if there was an easy way to solve this equation.
The equation is:
$$
\frac{30}{15+a} + \frac{30}{15-a} = 4.5
$$
The approach I tried was multiplying both sides by $(15+a)(15-a)$ to get rid of the fractions but I ended up with a really messy quadratic. I am sure there is a simpler way.
Thanks for any help :-)
Edit
Thanks for all the help, it seems there are many ways to do it, you just need a little clever thinking. Just in case anyone's interested I got this equation from an iq test question on a phone app. Which was:   
A motorboat, whose speed is 15km/hr in still water goes 30km downstream    
and comes back in a total of 4 hours 30 minutes. What is the speed    
of the stream's current?

Maybe I was barking up the wrong tree with this approach, the test was only supposed to take 5 minutes and had 10 questions. Is there a simpler way?
 A: Not that messy and a little algebra makes things really simple:
$$\frac{30}{15+a} + \frac{30}{15-a} = 4.5\iff450-30a+450+30a=\frac92(225-a^2)\iff$$
$$\iff1800=2025-9a^2\iff 9a^2=225\iff a^2=25\iff a=\pm 5$$
A: Rewrite the equation as
$$
30\left(\frac{1}{15+a}+\frac{1}{15-a}\right)=\frac{9}{2}
$$
or
$$
20\left(\frac{1}{15+a}+\frac{1}{15-a}\right)=3
$$
The term in parentheses can be developed to
$$
\frac{(15-a)+(15+a)}{(15+a)(15-a)}=\frac{30}{225-a^2}
$$
so your equation becomes
$$
20\frac{30}{225-a^2}=3
$$
or, better,
$$
\frac{200}{225-a^2}=1
$$
that is,
$$
200=225-a^2
$$
which should be easy.
Since there are two solutions, it is a quadratic. You can however predict that there will be no first degree term in the final equation because exchanging $a$ with $-a$ doesn't change the equation.
A: Not really.  This is a quadratic equation in disguise.  You can make the numbers smaller by defining $x=\frac a{15}$.  The equation becomes $\frac 2{1+x}+ \frac 2{1-x}=\frac 92$ and when you clear the denominators you get $4(1-x)+4(1+x)=9-9x^2, 0=1-9x^2$ which doesn't seem so messy.  Even without the substitution, you won't have a first power term, so can solve the equation just by taking a square root.
A: If you got a really messy quadratic, then you made a mistake.  Remember the identity: 
$$
(X+Y)(X-Y)=X^2-Y^2
$$
In particular:
$$
(15+a)(15-a)=225-a^2
$$
If you multiply both sides of your equation by $(15+a)(15-a)$ you get: 
\begin{align}
30(15-a)+30(15+a)&=4.5(225-a^2)\\
2\times30\times15&=4.5(225-a^2)\\
4.5a^2&=4.5\times225-2\times30\times15\\
a^2&=225-2\times30\times15/4.5=225-900/4.5=225-200=25
\end{align}
This is a quadratic with no linear term, so it should be particularly easy to solve!
A: There might be a clever method based on some type of substitution or other insight, but I just did the obvious (without a calculator, by the way):
$$\frac{30}{15 + a} \; + \; \frac{30}{15 - a} \; = \; \frac{9}{2}$$
Now divide both sides by $3$ to make the numbers smaller.
$$\frac{10}{15 + a} \; + \; \frac{10}{15 - a} \; = \; \frac{3}{2}$$
Now combine the fractions on the left side.
$$\frac{10(15 - a) \; + \; 10(15 + a)}{(15 + a)(15 - a)} \; = \; \frac{3}{2}$$
Now cross-multiply.
$$20(15 - a) \; + \; 20(15 + a) \; = \; 3(15+a)(15-a)$$
Now expand the left side.
$$(20)(15) \; - \; 20a \; + \; (20)(15) \; + \; 20a \; = \; 3(15+a)(15-a)$$
Now cancel the $\;20a$'s $\;$and use the fact that $\;(20)(15) + (20)(15) = (40)(15).$
$$(40)(15) \; = \; 3(15+a)(15-a)$$
Now divide both sides by $3$.
$$(40)(5) \; = \; (15+a)(15-a)$$
Now expand the right side. (Really easy because it's a factored form of a difference of squares.)
$$(40)(5) \; = \; 15^2 - a^2$$
Now isolate $a^2$.
$$a^2 \; = \; 15^2 - (40)(5)$$
Now factor the right hand side. The numbers are small enough that this isn't really needed, but if the numbers were larger, then the method I'll demonstrate could be useful.
$$a^2 \; = \; 3^2 \cdot 5^2 \; - \; 8 \cdot 5^2$$
Now factor out $5^2$.
$$a^2 \; = \; 5^2 (3^2 - 8)$$
Now do some arithmetic.
$$a^2 \; = \; 5^2(9 - 8)$$
$$a^2 \; = \; 5^2$$
Now take the square root of both sides.
$$a \; = \; \pm 5$$
A: Note this is like the reverse of a partial fraction equation. Working with the fractions $$\frac{30}{15-a}+\frac {30}{15+a} = 30\times \frac {30}{225-a^2}$$ And this makes it clear that you have an equation in $a^2$, rather than in $a$.
