Math worksheet containing ten quadratic equations designed for practice with the quadratic formula.

Let's solve each quadratic equation using the quadratic formula:

The quadratic formula is:

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

For a quadratic equation in the form:
$$
ax^2 + bx + c = 0
$$

We will go through each problem one by one, first rewriting them in standard form (if needed), then applying the formula.

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1) $ w^2 + 4w - 28 = 49 $

Step 1: Move all terms to one side:
$$
w^2 + 4w - 28 - 49 = 0 \Rightarrow w^2 + 4w - 77 = 0
$$

Now: $ a = 1, b = 4, c = -77 $

$$
w = \frac{-4 \pm \sqrt{4^2 - 4(1)(-77)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 308}}{2} = \frac{-4 \pm \sqrt{324}}{2}
$$

$$
\sqrt{324} = 18 \Rightarrow w = \frac{-4 \pm 18}{2}
$$

$$
w = \frac{-4 + 18}{2} = \frac{14}{2} = 7,\quad w = \frac{-4 - 18}{2} = \frac{-22}{2} = -11
$$

✔ Solutions: $ w = 7, -11 $

---

2) $ 12s^2 - 88s = 77 $

Bring all terms to one side:
$$
12s^2 - 88s - 77 = 0
$$

$ a = 12, b = -88, c = -77 $

$$
s = \frac{-(-88) \pm \sqrt{(-88)^2 - 4(12)(-77)}}{2(12)} = \frac{88 \pm \sqrt{7744 + 3696}}{24}
$$

$$
= \frac{88 \pm \sqrt{11440}}{24}
$$

Now simplify $ \sqrt{11440} $. Let's factor it:

$ 11440 = 16 \times 715 $ → $ \sqrt{11440} = \sqrt{16 \cdot 715} = 4\sqrt{715} $

So:
$$
s = \frac{88 \pm 4\sqrt{715}}{24} = \frac{4(22 \pm \sqrt{715})}{24} = \frac{22 \pm \sqrt{715}}{6}
$$

✔ Solutions: $ s = \frac{22 + \sqrt{715}}{6}, \frac{22 - \sqrt{715}}{6} $

(Note: This is exact form; decimal approx. would be about $ s \approx 8.1, -0.77 $)

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3) $ 6q^2 - 41q - 4 = 44 $

Move all to left:
$$
6q^2 - 41q - 4 - 44 = 0 \Rightarrow 6q^2 - 41q - 48 = 0
$$

$ a = 6, b = -41, c = -48 $

$$
q = \frac{-(-41) \pm \sqrt{(-41)^2 - 4(6)(-48)}}{2(6)} = \frac{41 \pm \sqrt{1681 + 1152}}{12}
$$

$$
= \frac{41 \pm \sqrt{2833}}{12}
$$

Check if $ \sqrt{2833} $ simplifies:
2833 ÷ 7 ≈ 404.7 → not divisible by small squares. So leave as is.

✔ Solutions: $ q = \frac{41 \pm \sqrt{2833}}{12} $

---

4) $ 8y^2 - 29y = -10 $

Bring to standard form:
$$
8y^2 - 29y + 10 = 0
$$

$ a = 8, b = -29, c = 10 $

$$
y = \frac{-(-29) \pm \sqrt{(-29)^2 - 4(8)(10)}}{2(8)} = \frac{29 \pm \sqrt{841 - 320}}{16} = \frac{29 \pm \sqrt{521}}{16}
$$

$ \sqrt{521} $ is irrational (not a perfect square)

✔ Solutions: $ y = \frac{29 \pm \sqrt{521}}{16} $

---

5) $ z^2 + 20z + 50 = 0 $

Already in standard form: $ a = 1, b = 20, c = 50 $

$$
z = \frac{-20 \pm \sqrt{20^2 - 4(1)(50)}}{2(1)} = \frac{-20 \pm \sqrt{400 - 200}}{2} = \frac{-20 \pm \sqrt{200}}{2}
$$

$ \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2} $

$$
z = \frac{-20 \pm 10\sqrt{2}}{2} = -10 \pm 5\sqrt{2}
$$

✔ Solutions: $ z = -10 + 5\sqrt{2},\ -10 - 5\sqrt{2} $

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6) $ g^2 - 29g - 80 = 0 $

$ a = 1, b = -29, c = -80 $

$$
g = \frac{-(-29) \pm \sqrt{(-29)^2 - 4(1)(-80)}}{2(1)} = \frac{29 \pm \sqrt{841 + 320}}{2} = \frac{29 \pm \sqrt{1161}}{2}
$$

Check $ \sqrt{1161} $:
1161 ÷ 3 = 387 → 387 ÷ 3 = 129 → 129 ÷ 3 = 43 → so $ 1161 = 3^3 \cdot 43 $, no perfect square factors.

✔ Solutions: $ g = \frac{29 \pm \sqrt{1161}}{2} $

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7) $ 24d^2 + 8d - 0 = 2 $

Wait — this says $ 24d^2 + 8d - 0 = 2 $, which is just:

$$
24d^2 + 8d = 2
$$

Bring to standard form:
$$
24d^2 + 8d - 2 = 0
$$

Divide entire equation by 2 for simplicity:
$$
12d^2 + 4d - 1 = 0
$$

Now: $ a = 12, b = 4, c = -1 $

$$
d = \frac{-4 \pm \sqrt{4^2 - 4(12)(-1)}}{2(12)} = \frac{-4 \pm \sqrt{16 + 48}}{24} = \frac{-4 \pm \sqrt{64}}{24}
$$

$ \sqrt{64} = 8 $

$$
d = \frac{-4 \pm 8}{24}
$$

$$
d = \frac{-4 + 8}{24} = \frac{4}{24} = \frac{1}{6},\quad d = \frac{-4 - 8}{24} = \frac{-12}{24} = -\frac{1}{2}
$$

✔ Solutions: $ d = \frac{1}{6},\ -\frac{1}{2} $

---

8) $ q^2 + 7q - 6 = 12 $

Bring all to left:
$$
q^2 + 7q - 6 - 12 = 0 \Rightarrow q^2 + 7q - 18 = 0
$$

$ a = 1, b = 7, c = -18 $

$$
q = \frac{-7 \pm \sqrt{7^2 - 4(1)(-18)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 72}}{2} = \frac{-7 \pm \sqrt{121}}{2}
$$

$ \sqrt{121} = 11 $

$$
q = \frac{-7 \pm 11}{2}
$$

$$
q = \frac{-7 + 11}{2} = \frac{4}{2} = 2,\quad q = \frac{-7 - 11}{2} = \frac{-18}{2} = -9
$$

✔ Solutions: $ q = 2, -9 $

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9) $ d^2 + 29d = -70 $

Bring to standard form:
$$
d^2 + 29d + 70 = 0
$$

$ a = 1, b = 29, c = 70 $

$$
d = \frac{-29 \pm \sqrt{29^2 - 4(1)(70)}}{2(1)} = \frac{-29 \pm \sqrt{841 - 280}}{2} = \frac{-29 \pm \sqrt{561}}{2}
$$

Check $ \sqrt{561} $:
561 ÷ 3 = 187 → 187 ÷ 11 = 17 → $ 561 = 3 \cdot 11 \cdot 17 $, no perfect squares.

✔ Solutions: $ d = \frac{-29 \pm \sqrt{561}}{2} $

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10) $ 10h^2 - 37h - 15 = 0 $

$ a = 10, b = -37, c = -15 $

$$
h = \frac{-(-37) \pm \sqrt{(-37)^2 - 4(10)(-15)}}{2(10)} = \frac{37 \pm \sqrt{1369 + 600}}{20} = \frac{37 \pm \sqrt{1969}}{20}
$$

Check $ \sqrt{1969} $:
1969 ÷ 11 = 179 → 179 is prime → no perfect square factors.

✔ Solutions: $ h = \frac{37 \pm \sqrt{1969}}{20} $

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✔ Final Answers Summary:

| Problem | Solutions |
|--------|-----------|
| 1) | $ w = 7, -11 $ |
| 2) | $ s = \frac{22 \pm \sqrt{715}}{6} $ |
| 3) | $ q = \frac{41 \pm \sqrt{2833}}{12} $ |
| 4) | $ y = \frac{29 \pm \sqrt{521}}{16} $ |
| 5) | $ z = -10 \pm 5\sqrt{2} $ |
| 6) | $ g = \frac{29 \pm \sqrt{1161}}{2} $ |
| 7) | $ d = \frac{1}{6},\ -\frac{1}{2} $ |
| 8) | $ q = 2, -9 $ |
| 9) | $ d = \frac{-29 \pm \sqrt{561}}{2} $ |
| 10) | $ h = \frac{37 \pm \sqrt{1969}}{20} $ |

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Let me know if you'd like decimal approximations or factoring where possible!