When substituting into the quadratic formula, keep careful track of signs, especially when $c$ is negative or when the quadratic starts with a negative $a$.
The Discriminant Predicts The Number Of Real Solutions
Inside the quadratic formula is the expression $$b^2-4ac$$
This is called the discriminant and is denoted by the Greek letter delta: $$\Delta=b^2-4ac$$
What Δ Tells You
If $\Delta>0$, there are 2 distinct real solutions.
If $\Delta=0$, there is 1 real solution (a repeated root).
If $\Delta<0$, there are no real solutions (the solutions are not real numbers).
Since the solutions to $ax^2+bx+c=0$ are exactly the x-intercepts of the graph $y=ax^2+bx+c$:
$\Delta>0$ means the parabola crosses the x-axis twice.
$\Delta=0$ means the parabola touches the x-axis once (tangent).
$\Delta<0$ means the parabola does not meet the x-axis.
Analogy
Think of $\Delta$ as a "preview" of the square root part of the quadratic formula.
If $\Delta$ is negative, you would need $\sqrt{\text{negative}}$, which is why there are no real solutions.
Using The Discriminant To Decide If A Quadratic Can Be Factorised
A quadratic with integer coefficients is factorisable into real linear factors only if it has real roots, so $\Delta$ must be non-negative.
If $\Delta<0$, it is not factorisable over the real numbers (and you will not find real roots).
If $\Delta\ge 0$, it might be factorisable, but you still need to check whether the roots are "nice" (for example, integers or simple fractions).
Note
In many school settings, "factorizable" often means factorizable with integer coefficients.
A quadratic can have $\Delta>0$ but still not factor neatly into integers (for example, if the roots involve radicals).
Example question
Solve $x^2=x+5$, and give exact answer.
Solution
Rearrange: $x^2-x-5=0$ so $a=1$, $b=-1$, $c=-5$.
Use the quadratic formula: $$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-5)}}{2(1)}}$$ $$=\frac{1\pm\sqrt{1+20}}{2}=$$ $$\frac{1\pm\sqrt{21}}{2}$$
So the solutions are $\displaystyle x=\frac{1+\sqrt{21}}{2}$ and $\displaystyle x=\frac{1-\sqrt{21}}{2}$.
Example question
Solve $x^2+0.6x-2.6=0$ (to 3 s.f.).
Solution
Here $a=1$, $b=0.6$, $c=-2.6$.
$$x=\frac{-0.6\pm\sqrt{0.6^2-4(1)(-2.6)}}{2}$$
$$=\frac{-0.6\pm\sqrt{0.36+10.4}}{2}$$
$$=\frac{-0.6\pm\sqrt{10.76}}{2}$$
Since $\sqrt{10.76}\approx 3.280$,
$x\approx \frac{-0.6+3.280}{2}=1.34$ (3 s.f.)
$x\approx \frac{-0.6-3.280}{2}=-1.94$ (3 s.f.)
Tip
When rounding, keep extra digits during calculation, then round only at the end.
Problem Solving With A Parameter (Using Δ)
Sometimes you are asked to choose values so the quadratic has a certain number of solutions.
This is where the discriminant is especially useful.
Example question
Find $k$ so that $x^2+kx+16=0$ has 1 solution.
Solution
A quadratic has 1 real (repeated) solution when $\Delta=0$.
Here $a=1$, $b=k$, $c=16$: $$\Delta=k^2-4(1)(16)=k^2-64$$
Set $\Delta=0$: $$k^2-64=0 \Rightarrow k^2=64 \Rightarrow k=\pm 8$$
So $k=8$ or $k=-8$.
Active recall
What does $\Delta<0$ tell you about x-intercepts?
If a quadratic has exactly one x-intercept, what must be true about $\Delta$?
Why is rearranging to "$=0$" essential before identifying $a$, $b$, and $c$?
Exam technique
In many problems you will decide your method based on the equation's form and the required answer.
Try factorisation first if the numbers are small and the quadratic looks "friendly".
Use completing the square when you want exact surd answers or when the quadratic is close to a perfect square.
Use the quadratic formula when factorisation is not obvious, or when the question requests decimals (e.g., 1 d.p. or 3 s.f.).
A quick method check: compute $\Delta=b^2-4ac$.
If $\Delta$ is not a perfect square (like 13, 21, 10.76), neat factorisation is unlikely.
If $\Delta$ is a perfect square (like 0, 1, 4, 9, 16, 25, ...), factorisation or simple roots are more likely.
Definition
Quadratic equation
An equation that can be written in the form $ax^2+bx+c=0$ where $a\neq 0$.
Definition
Completing the square
A method of rewriting a quadratic into a perfect-square form (plus or minus a constant), allowing solutions by taking square roots.
Definition
Discriminant (Δ)
The value $\Delta=b^2-4ac$ for $ax^2+bx+c=0$, which determines the number of real solutions.