Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$ 
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$

I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.

According to maple, $a=-3/4, b=21/4$ is one solution, and if $r$ is a solution of $2x^2-12x+15=0$ then $a=r/2, b=12-(11/2)r$ is another solution. As the quadratic here has two real zeros, there are three pairs $(a,b)$ of real numbers for your system.
Of course this is not an elegant solution, and even worse it relies on maple. But it seemed odd to me since when the equations are subtracted and the result solved for $b$, we get
$b=(26a^2-40a+21)/(8-6a)$, which may then be put into the first equation, so I would have expected a fourth degree equation for a.
A comment by @zyx prompted another look at this without use of maple. When the above expression for $b$ is put into the first equation, the factored equation for $a$ is
$$\frac{(4a+3)^2(8a^2-24a+15)}{4(3a-4)^2}=0.$$
The zero denominator at $a=4/3$ doesn't lead to a solution to both equations, and each zero of the linear and quadratic factors in the numerator leads to only one $b$ so that the pair $(a,b)$ satisfies both equations. So geometrically the two conics meet at exactly three points in the plane. Suspecting that one must be a point of tangency led me to compute the two gradients of the functions $f,g$ at the simplest of the points, namely the $(-3/4,21/4)$, and both gradients turned out to be parallel to the vector $(19,5)$. The other two intersections coming from the quadratic roots then (it would seem to me) are geometrically transverse intersections of the two conics. 
