The Quadratic Formula

Any quadratic equation of the form $a{x}^2+bx+c=0$ (where $a\ne 0$) can be solved using:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

This formula works on any quadratic — including those that cannot be factorised. The $\pm$ symbol produces two solution branches: one using $+\sqrt{b^2-4ac}$ and one using $-\sqrt{b^2-4ac}$. Whether real solutions exist depends on the discriminant $b^2-4ac$, covered below.

This is a Higher tier topic. On Foundation, algebraic quadratic solving is by factorising; approximate graphical solutions can also be assessed. Completing the square and the quadratic formula are Higher tier techniques.

The three constants $a$, $b$, and $c$ come directly from the equation written in the form $a{x}^2+bx+c=0$:

Symbol	Meaning	Example: $3x^2-5x+2=0$
$a$	Coefficient of $x^2$	$a=3$
$b$	Coefficient of $x$	$b=-5$
$c$	Constant term	$c=2$

You may be given the formula in some exam series, but you should not rely on that. You must know what each part means, identify $a$, $b$, $c$ correctly, substitute accurately, and simplify.

The equation must be in the form $a{x}^2+bx+c=0$ before you read off $a$, $b$, and $c$. If it is not in this form, rearrange first.

Rearrangement examples:

Original equation	Rearranged form	$a$, $b$, $c$
$x^2+5x=3$	$x^2+5x-3=0$	$a=1,b=5,c=-3$
$2x^2=7x-1$	$2x^2-7x+1=0$	$a=2,b=-7,c=1$
$x(x+4)=9$	$x^2+4x-9=0$	$a=1,b=4,c=-9$

Take particular care with the signs of $b$ and $c$. A sign error here flows through the entire calculation.

Worked example — Identify $a$, $b$, $c$ for $5-3x^2+x=0$.

Rearranging into standard form: $-3x^2+x+5=0$, so $a=-3$, $b=1$, $c=5$.

Alternatively multiply through by $-1$: $3x^2-x-5=0$, so $a=3$, $b=-1$, $c=-5$. Either form gives the same solutions.

Problem — Solve $2x^2+5x-3=0$, giving exact answers.

Read off: $a=2$, $b=5$, $c=-3$.

Substitute into the formula: $x=\frac{-5\pm \sqrt{5^2-4(2)(-3)}}{2(2)}=\frac{-5\pm \sqrt{25+24}}{4}=\frac{-5\pm \sqrt{49}}{4}=\frac{-5\pm 7}{4}$

Two solutions: $x=\frac{-5+7}{4}=\frac{2}{4}=\frac{1}{2}\text{or}x=\frac{-5-7}{4}=\frac{-12}{4}=-3$

Check: $2{\left(\frac{1}{2}\right)}^2+5\left(\frac{1}{2}\right)-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$ ✓ and $2(-3{)}^2+5(-3)-3=18-15-3=0$ ✓

Always substitute both answers back into the original equation to check. This catches arithmetic errors before they cost marks.

Problem — Solve $x^2-4x+1=0$, giving answers to 3 significant figures.

Read off: $a=1$, $b=-4$, $c=1$.

$x=\frac{4\pm \sqrt{(-4{)}^2-4(1)(1)}}{2(1)}=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}$

$x=\frac{4+\sqrt{12}}{2}\approx \frac{4+3.464}{2}\approx 3.73\text{or}x=\frac{4-\sqrt{12}}{2}\approx \frac{4-3.464}{2}\approx 0.268$

If exact answers are required, $\sqrt{12}=2\sqrt{3}$, so $x=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$.

The expression $b^2-4ac$ under the square root is called the discriminant. It tells you how many real solutions the equation has — before you complete the full calculation.

Discriminant value	Number of real solutions	Graph behaviour
$b^2-4ac>0$	Two distinct real solutions	Parabola crosses $x$-axis twice
$b^2-4ac=0$	One repeated solution	Parabola touches $x$-axis once
$b^2-4ac<0$	No real solutions	Parabola does not cross $x$-axis

Worked example — Without fully solving, determine how many solutions $x^2+2x+5=0$ has.

$a=1$, $b=2$, $c=5$. Discriminant $=4-20=-16<0$.

No real solutions. The parabola $y=x^2+2x+5$ sits entirely above the $x$-axis.

In exam questions, "show that the equation has no real solutions" means calculate the discriminant and show it is negative. You do not need to attempt the full formula.

The quadratic formula always works, but it is not always the fastest method. Choose based on what the question allows.

Method	When to use
Factorising	Equation factorises neatly (integers, small numbers) — fastest method
Completing the square	Question says "complete the square" or asks for the turning point
Quadratic formula	Equation does not factorise; decimal or surd answers needed; discriminant questions

Decision process: Try to spot factors first. If no integer factors exist after 30 seconds, switch to the formula. Never spend more than a minute attempting to factorise an equation that may not factorise.

1. Not rearranging to $a{x}^2+bx+c=0$ first

Reading $a$, $b$, $c$ from an equation that is not in standard form produces wrong values. For $x^2=5x-6$, students often use $b=5$ instead of $b=-5$.

2. Sign errors with $b$ and $-b$

The formula starts with $-b$. If $b=-4$, then $-b=4$. Writing $-(-4)=-4$ instead of $+4$ is a very common slip that invalidates both solutions.

3. Not including the entire numerator under $\pm$

$x=-5\pm \frac{\sqrt{49}}{4}$

is wrong. The $-5$ and $\pm \sqrt{49}$ must both be divided by $2a$:

$x=\frac{-5\pm \sqrt{49}}{4}$

4. Forgetting to calculate both solutions

The $\pm$ means two separate calculations. If the question asks for solutions (plural) and you give only one, you will lose marks.

5. Rounding too early

Keep $\sqrt{b^2-4ac}$ in full precision until the final step. Rounding $\sqrt{12}$ to $3.5$ before dividing introduces error. Use the full calculator display throughout.