<s>[INST] Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
 [/INST] For a lot of ways to solve quadratic equations, see Pat Ballew's manuscript Solving Quadratic Equations.
Here's a way to get the quadratic formula if you partially remember it. (I don't know off-hand if this is one of the methods in Pat Ballew's list.)
We remember that the quadratic formula gives us results that can be put into the form $A \pm \sqrt{B}.$ Note that we don't need to incorporate a constant factor with the square root, since we can bring the square of any such factor inside the square root: $7\sqrt{3}$ is equal to $\sqrt{7^2 \cdot 3}.$
Thus, we know the answer will be
$$x \; = \; A \pm \sqrt{B}$$
Isolating the radical and squaring gives:
$$(x - A)^2 \; = \; \left( \pm \sqrt{B} \right)^2$$
$$(x - A)^2 \; = \; B$$
$$(x - A)^2 - B \; = \; 0$$
$$ x^2 - 2Ax + (A^2 - B) \; = \; 0 $$
Now let's solve $x^2 + bx + c = 0.$ [By dividing through by the coefficient of $x^2,$ we may assume the coefficient of $x^2$ is $1.]$
Comparing the coefficients of $x^2 + bx + c = 0$ with the coefficients of the last displayed equation, we see that $x = A \pm \sqrt{B},$ where $-2A = b$ and $(A^2 - B) = c.$ Solving for $A$ and $B,$ we get $A = -\frac{b}{2}$ and $B = \frac{b^2}{4} - c,$ and now we have the solution for $x$ in terms of the equation's coefficients $b$ and $c.$
</s>