1. What a quadratic is
An equation that can be written in the form $ax^2 + bx + c = 0$, where $a$, $b$, $c$ are constants and $a \neq 0$. (If $a = 0$, the equation is no longer quadratic — it's linear.) The exponent $2$ is the defining feature; "quadratic" comes from the Latin quadratus, meaning "square."
Three coefficients: $a$ (leading), $b$ (middle), $c$ (constant). Each can be positive, negative, or zero, with the lone constraint that $a \neq 0$. The form is called standard form.
Most quadratic equations have either two real solutions, one real solution, or zero real solutions — never three, never infinitely many. This is true of any polynomial equation of degree $n$: it has at most $n$ real solutions, and exactly $n$ complex solutions counted with multiplicity (the Fundamental Theorem of Algebra).
2. Solving by factoring
If a quadratic factors nicely, the fastest path to the solutions is the zero-product property:
If $A \cdot B = 0$, then $A = 0$ or $B = 0$ (or both).
This property is special to multiplication by zero, and it's what makes factoring useful. If we can rewrite $ax^2 + bx + c$ as a product $(x - r_1)(x - r_2)$ multiplied by some constant, then the equation $= 0$ is satisfied precisely when one of the factors is zero — that is, when $x = r_1$ or $x = r_2$.
Example: solve $x^2 - 5x + 6 = 0$.
Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$. (Two numbers multiplying to $6$, adding to $-5$: $-2$ and $-3$.)
Apply zero product property: $(x - 2)(x - 3) = 0 \Rightarrow x - 2 = 0$ or $x - 3 = 0$, so $x = 2$ or $x = 3$.
Both work — substitute back: $4 - 10 + 6 = 0$ ✓ and $9 - 15 + 6 = 0$ ✓.
$x^2 - 2 = 0$ has solutions $x = \pm\sqrt{2}$ — irrational, not integer. Factoring won't find them by inspection. When factoring fails, switch to completing the square or the quadratic formula.
3. Completing the square
The technique that works even when factoring doesn't — and the engine behind the quadratic formula. The idea: rewrite the quadratic as a perfect square plus a constant, then isolate the square and take the square root.
Solve $x^2 + 6x - 7 = 0$:
Step 1. Move the constant to the other side:
$$ x^2 + 6x = 7. $$Step 2. Take half the middle coefficient ($\tfrac{6}{2} = 3$), square it ($9$), and add to both sides:
$$ x^2 + 6x + 9 = 7 + 9 = 16. $$Step 3. The left side is now a perfect square trinomial:
$$ (x + 3)^2 = 16. $$Step 4. Take the square root of both sides — remembering both signs:
$$ x + 3 = \pm 4. $$Step 5. Solve for $x$:
$$ x = -3 + 4 = 1 \quad \text{or} \quad x = -3 - 4 = -7. $$The trick is in step 2: completing the square literally means turning a partial $x^2 + bx$ into a complete perfect-square trinomial by adding the right constant. The constant is $\left(\tfrac{b}{2}\right)^2$.
A perfect-square trinomial has the form $(x + k)^2 = x^2 + 2kx + k^2$. If you want $x^2 + bx$ to fit this template, the middle term forces $2k = b$, so $k = \tfrac{b}{2}$, and the constant you need to add is $k^2 = \left(\tfrac{b}{2}\right)^2$. Half the coefficient, then square.
4. The quadratic formula
Apply completing the square to the general quadratic $ax^2 + bx + c = 0$ — once — and you get a formula that solves any quadratic at all.
Starting from $ax^2 + bx + c = 0$:
Divide everything by $a$:
$$ x^2 + \frac{b}{a}x + \frac{c}{a} = 0. $$Move the constant:
$$ x^2 + \frac{b}{a}x = -\frac{c}{a}. $$Complete the square. Half of $\tfrac{b}{a}$ is $\tfrac{b}{2a}$; its square is $\tfrac{b^2}{4a^2}$. Add to both sides:
$$ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}. $$The left side is now $\left(x + \tfrac{b}{2a}\right)^2$. The right side simplifies to $\tfrac{b^2 - 4ac}{4a^2}$:
$$ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. $$Take the square root:
$$ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}. $$Solve for $x$:
$$ \boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}. $$That's the quadratic formula. Memorize it, but more importantly remember it came out of completing the square — so if you ever forget the formula, you can rederive it in a minute.
5. The discriminant
The expression under the square root, $b^2 - 4ac$, is called the discriminant. Its sign tells you what kind of solutions the equation has, without your ever computing them:
| Discriminant $b^2 - 4ac$ | Number of real solutions | What the graph does |
|---|---|---|
| Positive ($> 0$) | Two distinct real solutions | Parabola crosses the x-axis at two points. |
| Zero ($= 0$) | One real solution (a "double root") | Parabola just touches the x-axis once at its vertex. |
| Negative ($< 0$) | No real solutions (two complex) | Parabola never crosses the x-axis. |
The discriminant lets you preview the answer. If a textbook problem yields a negative discriminant, either the problem expects complex solutions or you've made an arithmetic mistake.