Solving quadratic equations

Solving a quadratic equation means finding every value of $x$ that makes it true.
The key idea is the zero-product rule: if two things multiply to $0$, then at least one of them must be $0$.
So we rewrite the equation as (something)(something) $= 0$, then solve each piece.

The most common method is factoring, and it follows a set recipe.
For $ax^2 + bx + c = 0$, first multiply $a$ and $c$ together.
Then find two numbers that multiply to $a \cdot c$ and add to $b$.
Use those numbers to split the middle term, group, and pull out the common factors.
Take $x^2 + 5x + 6 = 0$, where $a = 1$, $b = 5$, $c = 6$, so $a \cdot c = 6$.
The pairs that multiply to $6$ are $1$ and $6$, or $2$ and $3$; the pair that adds to $b = 5$ is $2$ and $3$.
So it factors as $(x + 2)(x + 3) = 0$, giving $x = -2$ or $x = -3$.

When a quadratic will not factor nicely, use the quadratic formula: Here $a$, $b$, and $c$ are the numbers in $ax^2 + bx + c = 0$.
For $x^2 - x - 2 = 0$, we have $a = 1$, $b = -1$, $c = -2$.
Plugging in gives $x = \frac{1 \pm 3}{2}$, so $x = 2$ or $x = -1$.

If the equation is a perfect square set equal to a number, take the square root of both sides.
For $(2x + 3)^2 = 25$, take the square root to get $2x + 3 = \pm 5$.
That splits into $2x + 3 = 5$ and $2x + 3 = -5$.
Solving each gives $x = 1$ or $x = -4$.
Always keep the $\pm$: a square root gives both a positive and a negative answer.

You can also find the solutions from a graph.
Plot the parabola (a graphing tool like Desmos works well).
Wherever it crosses the $x$-axis, that $x$-value is a solution.

One caution: never divide both sides by $x$ to solve, because $x$ might be $0$.
Instead, move everything to one side and factor out $x$.

Solve $x^2 + 7x + 12 = 0$ by factoring.
Here $a = 1$, $b = 7$, $c = 12$, so $a \cdot c = 12$.
Find two numbers that multiply to $12$ and add to $7$: those are $3$ and $4$.
So $(x + 3)(x + 4) = 0$, giving $x = -3$ or $x = -4$.

Solve $2q^2 - 9q - 5 = 0$ by factoring.
Here $a = 2$, $b = -9$, $c = -5$, so $a \cdot c = -10$.
Find two numbers that multiply to $-10$ and add to $-9$: those are $-10$ and $1$.
Split the middle term: $2q^2 - 10q + q - 5 = 0$.
Group and factor: $2q(q - 5) + 1(q - 5) = 0$, so $(q - 5)(2q + 1) = 0$.
This gives $q = 5$ or $q = -\frac{1}{2}$.

Solve $x^2 - 4x - 5 = 0$ using the quadratic formula.
Here $a = 1$, $b = -4$, and $c = -5$.
So $x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2}$.
This gives $x = 5$ or $x = -1$.