One way to solve quadratics is by factoring. The catch with factoring is that you’ll need to have zero on one side of the equation.
\[(x+2)(x+4) = 10 \qquad (x+2)(x+4) = 0\]The equation on the left, despite being factored, isn’t really solvable the way it’s written. You would need to find some value of $x$ that when added to both 2 and 4, the two sums would multiply to 10. The one on the right, however, is much easier since you only need one of the factors to be zero.
There is a way to solve quadratics without requiring a 0 on one side, but it has another requirement.
\[\begin{align*} x^2 + 6x + 9 &= 25 \\ (x+3)(x+3) &= 25 \\ (x + 3)^2 &= 25 \\ x + 3 &= \pm 5 \\ x &= -3 \pm 5 \end{align*}\]We end up with the solutions $x=2$ and $x=-8$. So, we could solve it despite not having a zero. The catch is that the quadratic itself was a perfect square trinomial, the result of a binomial being squared. This made it very convenient to solve since once you got it back to a squared binomial, you can square root both sides.
Let’s take this idea and apply it to our very first equation, the one where solving by factoring didn’t work. Here it is in its unfactored form.
\[x^2 + 6x + 8 = 10\]The left side of the equation is not a perfect square trinomial. For it to be one, that 8 would have to be a 9. Well, let’s add 1 to both sides of the equation to get that done.
\[x^2 + 6x + 9 = 11\]Now we can factor, square root and solve.
\[\begin{align*} x^2 + 6x + 9 &= 11 \\ (x+3)^2 &= 11 \\ x + 3 &= \pm \sqrt{11} \\ x &= -3 \pm \sqrt{11} \end{align*}\]This strategy is known as completing the square and the trick to it is finding the correct $c$ value needed to make your quadratic a perfect square. Given that $a$ is 1, then your $c$ is just half of $b$ squared. All the trinomials below are perfect squares, so they have that property.
\[x^2 + 4x + 4 \qquad x^2 + 10x + 25 \qquad x^2 + 3x + \frac{9}{4}\]It’s trickier when $a$ is something other than 1, and usually the easiest way to deal with those is to factor out the $a$, then work with your new $b$.
\[\begin{align*} -2x^2 + 12x &= 1 \\ -2(x^2 - 6x) &= 1 \\ -2(x^2 - 6x + \mathbin{\color{red}{+9}}) &= 1 \color{red} + (-2)(9) \end{align*}\]Before moving on with this one, note that the right side also got a 9, but it was multiplied by −2. Since the 9 we added on the left is inside parentheses that are being multiplied by −2, we have to make sure we do the same on the right.
\[\begin{align*} -2(x^2 - 6x + 9) &= 1 + (-2)(9) \\ -2(x-3)^2 &= -17 \\ (x-3)^2 &= \frac{17}{2} \\ x-3 &= \pm \sqrt{\frac{17}{2}} \\ x &= 3 \pm \sqrt{\frac{17}{2}} \end{align*}\]