How to Solve Quadratic Equations Step by Step (With Worked Examples)

A quadratic equation is any equation you can rearrange into the form $a{x}^2$ + bx + $c=0$ with a ≠ 0. That single shape c/s a huge slice of high school and early college math: projectile heights, area problems, optimization, and most of what shows up on standardized tests. The hard part is rarely the formula — it is choosing the right method and not making sign mistakes along the way.

This post walks through the three methods you actually need: factoring, completing the square, and the quadratic formula. For each one, we show the steps, work through a real example, and call out the places where students lose points without realizing it.

1. Factoring: fastest when the numbers cooperate

Factoring works when $a{x}^2$ + bx + c can be written as a product of two linear pieces, like (x - p)(x - q). The roots are then $x=p$ and $x=q$. The trick for the common case a $=1$ is to find two numbers that multiply to c and add to b.

Worked example

Solve $x^2-5x+6=0$. You need two numbers whose product is 6 and whose sum is -5. The pair (-2, -3) works: -2 · $-3=6$ and $-2+-3=-5$. $Sotheequationfactorsas(x-2)(x-3)=0$, giving $x=2$ or $x=3$.

When a ≠ 1, multiply a · c first, find a pair that multiplies to ac and adds to b, then split the middle term. It is the same idea, just one extra step.

2. Completing the square: the method that explains the formula

Completing the square is slower than factoring but it always works, and it is the move that makes the quadratic formula obvious. The idea: turn $a{x}^2$ + bx + $c=0$ into something of the for m $(x+h{)}^2$ $=k$, then take the square root of both sides.

Worked example

Solve $x^2+6x-7=0$. Move the constant: $x^2+6x=7$. Take half of the $x-coefficient(\frac{6}{2}=3),squareit(9)$, and add it to both sides: $x^2+6x+9=16$. The left side is now a perfect square: $(x+3{)}^2$ $=16$. Take the square root: $x+3=$±4. $Sox=1$ or $x=-7.$

3. The quadratic formula: the universal tool

The $quadraticformula,x=(-b$± $\sqrt{b^2-4ac})/2a$ is just completing the square done once for the general case. It always works, even when factoring is impossible. The expression under the square root, $b^2$ - 4ac, is the discriminant — and it tells you in advance how many real roots to expect.

• If $b^2$ - 4ac > 0: two distinct real roots.
• If $b^2$ - 4ac = 0: one repeated real root (the parabola is tangent to the x-axis).
• If $b^2$ - 4ac < 0: no real roots — two complex roots that come as a conjugate pair.

Worked example

Solve $2x^2-4x-3=0$. Here a $=2,b=-4,c=-3$. The discriminant is $(-4{)}^2-4$·2·$(-3)=16+24=40,so$ two real roots. Plug in: $x=(4$± $\surd 40)/4=(4$± $2\surd 10)/4=1$± $\surd \frac{10}{2}$. Leave it in exact form unless you are explicitly asked for a decimal.

How to choose a method

1. Try factoring first if the coefficients are small integers — it is the fastest when it works.
2. If factoring fails or feels forced, fall back to the quadratic formula. Always.
3. Use completing the square when the question explicitly asks for vertex form, or when you need to derive something rather than just solve it.

Sanity-checking your answer

Plug each root back into the original equation. If $2x^2-4x-3=0$ and you got $x=1+\surd \frac{10}{2}$, evaluating $2(1+\surd \frac{10}{2}{)}^2-4(1+\surd \frac{10}{2})-3$ must give zero. This catches almost every arithmetic slip.

If you have a quadratic in front of you and you would rather see the steps narrated out loud as someone writes them on a whiteboard, paste it into Solvequill — the explanation video shows the working in real time, names the method, and walks through the substitution. It is genuinely useful when you have a homework deadline at midnight.