Quadratic equations for math practice.

Let's solve each quadratic equation step by step. We'll use factoring where possible, and if factoring is difficult, we’ll use the quadratic formula:

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

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1. $ 2x^2 - 2x - 24 = 0 $

We can factor this.

First, factor out the GCF (if any):
No common factor in all terms.

Now look for two numbers that multiply to $ 2 \times (-24) = -48 $ and add to $-2$.

Try: $6$ and $-8$:
$6 \times (-8) = -48$, $6 + (-8) = -2$ ✔

So split the middle term:
$$
2x^2 + 6x - 8x - 24 = 0
$$
Group:
$$
(2x^2 + 6x) - (8x + 24) = 0 \\
2x(x + 3) - 8(x + 3) = 0 \\
(2x - 8)(x + 3) = 0
$$

Wait — actually, let's check:
$2x(x + 3) = 2x^2 + 6x$, and $-8(x + 3) = -8x -24$, so total: $2x^2 -2x -24$. Correct.

But we can simplify further:
$$
(2x - 8)(x + 3) = 0
\Rightarrow 2(x - 4)(x + 3) = 0
$$

So solutions:
$$
x = 4 \quad \text{or} \quad x = -3
$$

✔ Answer: $ x = 4 $ or $ x = -3 $

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2. $ x^2 + 10x + 16 = 0 $

Find two numbers that multiply to $16$ and add to $10$:
Try: $2$ and $8$: $2 \times 8 = 16$, $2 + 8 = 10$ ✔

So:
$$
(x + 2)(x + 8) = 0
\Rightarrow x = -2 \quad \text{or} \quad x = -8
$$

✔ Answer: $ x = -2 $ or $ x = -8 $

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3. $ 4x^2 + 15x + 9 = 0 $

Look for two numbers that multiply to $4 \times 9 = 36$ and add to $15$.

Try: $12$ and $3$: $12 \times 3 = 36$, $12 + 3 = 15$ ✔

Split middle term:
$$
4x^2 + 12x + 3x + 9 = 0 \\
(4x^2 + 12x) + (3x + 9) = 0 \\
4x(x + 3) + 3(x + 3) = 0 \\
(4x + 3)(x + 3) = 0
$$

Solutions:
$$
x = -\frac{3}{4} \quad \text{or} \quad x = -3
$$

✔ Answer: $ x = -\frac{3}{4} $ or $ x = -3 $

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4. $ x^2 - 13x + 42 = 0 $

Find two numbers that multiply to $42$ and add to $-13$.

Try: $-6$ and $-7$: $(-6)(-7) = 42$, $-6 + (-7) = -13$ ✔

So:
$$
(x - 6)(x - 7) = 0
\Rightarrow x = 6 \quad \text{or} \quad x = 7
$$

✔ Answer: $ x = 6 $ or $ x = 7 $

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5. $ 6x^2 - x - 40 = 0 $

Multiply $6 \times (-40) = -240$. Find two numbers that multiply to $-240$ and add to $-1$.

Try: $-16$ and $15$: $-16 \times 15 = -240$, $-16 + 15 = -1$ ✔

Split:
$$
6x^2 -16x + 15x - 40 = 0 \\
(6x^2 -16x) + (15x - 40) = 0 \\
2x(3x - 8) + 5(3x - 8) = 0 \quad \text{(Wait, let's check)}
$$

Actually:
$6x^2 -16x = 2x(3x - 8)$? $2x \cdot 3x = 6x^2$, $2x \cdot (-8) = -16x$ → yes
$15x - 40 = 5(3x - 8)$? $5 \cdot 3x = 15x$, $5 \cdot (-8) = -40$ → yes!

So:
$$
(2x + 5)(3x - 8) = 0
$$

Solutions:
$$
x = -\frac{5}{2} \quad \text{or} \quad x = \frac{8}{3}
$$

✔ Answer: $ x = -\frac{5}{2} $ or $ x = \frac{8}{3} $

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6. $ 2x^2 + 13x + 15 = 0 $

Multiply $2 \times 15 = 30$. Find two numbers that multiply to $30$ and add to $13$.

Try: $10$ and $3$: $10 \times 3 = 30$, $10 + 3 = 13$ ✔

Split:
$$
2x^2 + 10x + 3x + 15 = 0 \\
(2x^2 + 10x) + (3x + 15) = 0 \\
2x(x + 5) + 3(x + 5) = 0 \\
(2x + 3)(x + 5) = 0
$$

Solutions:
$$
x = -\frac{3}{2} \quad \text{or} \quad x = -5
$$

✔ Answer: $ x = -\frac{3}{2} $ or $ x = -5 $

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✔ Final Answers:

| Equation | Solutions |
|--------|-----------|
| $2x^2 - 2x - 24 = 0$ | $x = 4$, $x = -3$ |
| $x^2 + 10x + 16 = 0$ | $x = -2$, $x = -8$ |
| $4x^2 + 15x + 9 = 0$ | $x = -\frac{3}{4}$, $x = -3$ |
| $x^2 - 13x + 42 = 0$ | $x = 6$, $x = 7$ |
| $6x^2 - x - 40 = 0$ | $x = -\frac{5}{2}$, $x = \frac{8}{3}$ |
| $2x^2 + 13x + 15 = 0$ | $x = -\frac{3}{2}$, $x = -5$ |

Let me know if you'd like these shown with graphs or steps using the quadratic formula!