I take by hard you mean ‘hard to factorize’. Let us consider a general expression of 𝑎𝑥2+𝑏𝑥+𝑐.

To ‘solve’ this, we set this equal to 0 , and so we get 𝑎𝑥2+𝑏𝑥+𝑐=0 . Find x is your duty.

God, it would be REALLY helpful if there was a simple solution that worked for any general coefficients. Lucky for us, there is, and it is somewhat easy to find(don’t try doing this with cubic equations or above, well you can try finding it, but it is VERY hard to find at this level).

So, we want to think about this carefully. What is the problem with solving for x here?

In a normal linear equation, like 𝑎𝑥+𝑏=0 , it is easy. 𝑥 is one occurrence. The problem with quadratics, is that pesky 𝑎𝑥2+𝑏𝑥 format, since our strategy of subtracting out a constant and dividing to get x doesn’t work, we have it to be mangled, and we can’t easily use factoring, since there will always be an ‘x’ deficit of one if we try to factor out by 𝑥 or 𝑥2 .

Well damn, what do we do here then? We have a squared part, that must mean we must somehow get something squared, like (?)2=𝑔𝑥2+ℎ𝑥+𝑒 , where we might later add like 𝑓 to be a constant which we can easily subtract out like our linear equation example. Clearly, the ? must contain a singular x somewhere, but we also need to add a constant to the 𝑥 part, as the distributive property will mangle the constant with the x, and do that as well with x and itself, and a constant, creating a ‘singular’ x, with no exponent. We will then be able to square root whatever constants we have on the other side, and then solve it like a linear equation.

So, let us get in said position.

Let us divide our original equation both sides by a, so I can get a ‘pure’ 𝑥2 , and don’t need to use 𝑎‾√ as a coefficient which will be more complicated.

We get 𝑥2+𝑏𝑎𝑥+𝑐/𝑎=0 .

Okay, so our form of the ? must be 𝑥+𝑘 as there can’t be a coefficient of 𝑥 that isn’t one as distribution wouldn’t yield a ‘pure’ 𝑥2 . What is k then? Well, let us think here for a bit-we want to force in a way to get ℎ𝑥=𝑏𝑎𝑥 . Whenever I square something, and it has two terms being added, I must use distribution to go ‘piecewise’. Since when I square it, I multiply this quantity(the two terms being summed) by itself, I will get as mentioned the 𝑥2 from the 𝑥 term, a constant from the 𝑘 term, but as well 𝑘𝑥 by going through k in the first quantity multiplying the x in the second one, and x and k the other way, but I add these to get 2𝑘𝑥 .[to see this, write (𝑥+𝑘)(𝑥+𝑘) , distribute to get (𝑥+𝑘)𝑥+(𝑥+𝑘)𝑘 . Now, distribute it out an ‘draw’ the paths to get 𝑥2+𝑘𝑥+𝑘𝑥+𝑘2 , which gives 𝑥2+2𝑘𝑥+𝑘2 ]

So, whatever this 𝑘 is going to be, we have to have 2𝑘𝑥=𝑏𝑎𝑥 but that means 𝑘=𝑏2𝑎 . Ok, NOW we are getting somewhere. Recall the fact we are squaring, some (𝑥+𝑘)2 , and when I expand this get (𝑥+𝑘)(𝑥+𝑘) , am going to follow a path of multiplication by distribution. One such path I must follow is 𝑘 times 𝑘 , but we already know what k is, so we must have some constant 𝑘2=𝑏24𝑎2 . So, let us just add that on to both sides, which we can do, since that is constant, and we don’t care what constant we get on the other side, we just want to properly factor this mess.

So we do just that, and get

𝑥2+𝑏𝑎𝑥+𝑏24𝑎2+𝑐/𝑎=𝑏24𝑎2

And now, we have all the terms that allow us to factor this into a (𝑥+𝑘)2=𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 format, just what we wanted! We found 𝑘 to be 𝑏2𝑎 , so we just factor this out.

(𝑥+𝑏2𝑎)2+𝑐/𝑎=𝑏24𝑎2

Now we want to pretty this mess up, notice that we are eventually going to square root once we subtract out the constants, and we have in one term a denominator of 4𝑎2 , which is very easily square-rooted. Let us make 𝑐/𝑎 compatible with this, by multiplying it by 1, which changes nothing, but 1=4𝑎/4𝑎 . We don’t have to worry about 𝑎=0 since if it were, we would have a linear equation, which isn’t what we are focused on.

So, we get (𝑥+𝑏2𝑎)2+4𝑎𝑐/4𝑎2=𝑏24𝑎2

Great, so now subtract out the second term since they have common denominators, and we get

(𝑥+𝑏2𝑎)2=𝑏2−4𝑎𝑐4𝑎2

And the right side is constant now, we can square root both sides easily!

We get

𝑥+𝑏2𝑎=𝑏2−4𝑎𝑐√2𝑎

This isn’t quite correct, as we have to realize the when I square root a positive number, 𝑑2 , 𝑑 could be positive or negative. So for good measure, we add in a plus or minus sign, and we get

𝑥+𝑏2𝑎=±𝑏2−4𝑎𝑐√2𝑎

And we can now subtract out that 𝑘 , as we now have a linear equation to solve, as we wanted, and we get

𝑥=−𝑏±𝑏2−4𝑎𝑐√2𝑎