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I'm wondering how I solve this kind of equation.
I'm (re)learning 2nd degree equations, and I would have no problem solving $25x^2 - 9$ or $25x^2 - 34x + 9$, for example, but these ones with letters and exponents in the last term are not what I'm used to.
Suppose that the $25x^2 - 9y^2 = 0$ (I took the expression you gave in the title and made an example equation). Here we observe that $25$ and $9$ are perfect squares, so we can rewrite as $(y)^2 = (\frac{5x}{3})^2$. So we see that $y = \pm \dfrac{5x}{3}$. More often than not, there will be a pattern like this. A very useful tool, although simple, is the quadratic formula. In quadratics with two variables, for example $16x^2 + 2xy + 15y^2 = 0$ it is useful to find the value of $x$ or $y$ in terms of the other variable by using the quadratic formula. Another trick is to split up a sum into products and then calling them factors. For example in $m^3 - n^3 = 21$ I can apply the difference of cubes formula and call the subsequent terms factors of $21$ and solve simultaneous equations for each pair of factors.
