When solving quadratics, we have a few strategies. Factoring, and grouping when $a\neq1$, typically only work when solutions are integers, and not very large ones. Completing the square can help with the others, especially when the solutions are irrational or complex numbers.
When you use completing the square, you might have noticed the path to each solution felt the same. Well, if we apply those steps to our general quadratic formula of $ax^2 + bx + c$, we can develop a formula to cut out the steps needed to solve.
\[\begin{align*} ax^2 + bx + c &= 0 \\ ax^2 + bx &= -c \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} \\ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 &= -\frac{c}{a}+ \left(\frac{b}{2a}\right)^2 \\ \left(x + \frac{b}{2a}\right)^2 &= -\frac{c}{a}+ \frac{b^2}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 &= -\frac{4ac}{4a^2}+ \frac{b^2}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}\\ x + \frac{b}{2a} &= \pm\sqrt{\frac{b^2-4ac}{4a^2}}\\ x &= -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} \\[1em] x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \end{align*}\]Bit of a journey, but the last line is the quadratic formula. Given any quadratic written in the form $ax^2+bx+c=0$, you can use that to get your solutions.
The part of the quadratic formula under the radical sign can be helpful if you just want to know how many solutions a a quadratic has. Remember than quadratics make parabolas when graphed, so they can cross the $x$-axis twice, never, or even just once.
That last situation, where a parabola only crosses once, we haven’t discussed much, but an example is below.
\[\begin{align*} x^2 + 2x + 1 &= 0 \\ (x + 1)^2 &= 0 \\ x &= -1 \pm \sqrt{0}\\ x &= -1 \end{align*}\]We have a perfect square trinomial set equal to 0, so we end up with the same solution twice. This makes a parabola where the vertex is on the $x$-axis.
If you want to know how many solutions a quadratic has, not what they are just how many, then use the part under the radical $b^2-4ac$, which is the discriminant.