The Quadratic Formula: How and Why It Works

Compute the discriminant first. It is a perfect square, so the solutions are rational.

$x = -3$ or $x = -4$

$a = 1$, $b = 7$, $c = 12$.

$b^2 - 4ac = 49 - 48 = 1$

$x = \tfrac{-7 \pm 1}{2}$, giving $x = -3$ or $x = -4$.

Check: $(-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0$ ✓

$x = \tfrac{5}{3}$ or $x = -1$

$a = 3$, $b = -2$, $c = -5$.

$b^2 - 4ac = 4 + 60 = 64$

$x = \tfrac{2 \pm 8}{6}$, giving $x = \tfrac{10}{6} = \tfrac{5}{3}$ or $x = \tfrac{-6}{6} = -1$.

None.

$b^2 - 4ac = 4 - 20 = -16$, which is negative, so there are no real solutions. The parabola sits entirely above the $x$-axis.

Compute the discriminant before doing anything else.

$x = 3$, a single repeated solution.

$a = 1$, $b = -6$, $c = 9$.

$b^2 - 4ac = 36 - 36 = 0$, so there is exactly one solution.

$x = \tfrac{6 \pm 0}{2} = 3$.

The parabola touches the $x$-axis at $(3, 0)$ without crossing it.

The discriminant is not a perfect square, so a square root will stay in the answer.

$x = -2 + \sqrt{3}$ or $x = -2 - \sqrt{3}$

$b^2 - 4ac = 16 - 4 = 12$.

$x = \tfrac{-4 \pm \sqrt{12}}{2}$. Since $\sqrt{12} = 2\sqrt{3}$, this becomes $\tfrac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}$.

Both parts of the numerator were divided by $2$, not only the root.

It is not equal to zero yet.

$x = 2$ or $x = \tfrac{1}{3}$

Move everything to one side: $3x^2 - 7x + 2 = 0$, so $a = 3$, $b = -7$, $c = 2$.

$b^2 - 4ac = 49 - 24 = 25$.

$x = \tfrac{7 \pm 5}{6}$, giving $x = 2$ or $x = \tfrac{2}{6} = \tfrac{1}{3}$.

One.

$b^2 - 4ac = 144 - 4(4)(9) = 144 - 144 = 0$, so there is exactly one repeated solution.

The discriminant answers “how many” without any need to answer “which”.

Dividing through by $-2$ first makes the arithmetic easier and changes nothing.

$x = 3$ or $x = -1$

Divide every term by $-2$: $x^2 - 2x - 3 = 0$.

$b^2 - 4ac = 4 + 12 = 16$, so $x = \tfrac{2 \pm 4}{2}$, giving $x = 3$ or $x = -1$.

Using $a = -2$, $b = 4$, $c = 6$ directly gives the same pair, with a division by $-4$ along the way.

Ground level means $h = 0$. Two answers will appear — decide what each one means.

After $4$ seconds.

Set $h = 0$: $-5t^2 + 20t = 0$, so $a = -5$, $b = 20$, $c = 0$.

$b^2 - 4ac = 400 - 0 = 400$, and $x = \tfrac{-20 \pm 20}{-10}$, giving $t = 0$ or $t = 4$.

Both are real, and both mean something: $t = 0$ is the moment of the throw, when the ball was also at ground level, and $t = 4$ is the landing. The question asks for the landing, so the answer is $4$ seconds.

Check the sign of $c$.

She used $c = 5$ instead of $c = -5$. The solutions are $x = 5$ and $x = -1$.

With $c = -5$, the calculation is $b^2 - 4ac = 16 - 4(1)(-5) = 16 + 20 = 36$.

$x = \tfrac{4 \pm 6}{2}$, giving $x = 5$ or $x = -1$.

Checking: $5^2 - 4(5) - 5 = 25 - 20 - 5 = 0$ ✓

A dropped minus sign in $c$ flips $-4ac$ from adding to subtracting, which is why it so often turns a solvable equation into an apparently unsolvable one.