Quadratic equations worksheet for solving for x.

We are given a worksheet titled “QUADRATIC EQUATIONS” with 12 quadratic equations to solve for x. The goal is to find the solution(s) to each equation.

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General Strategy:

Most of these equations can be solved by:
1. Moving all terms to one side to set the equation equal to zero.
2. Factoring, if possible.
3. If factoring is not easy, use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
for an equation in the form \( ax^2 + bx + c = 0 \).

Let’s solve each equation step-by-step.

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## ✔ Equation 1: \( x^2 + 4x + 3 = 0 \)

Factor:
\[
(x + 1)(x + 3) = 0
\]
Solutions:
\[
x = -1, \quad x = -3
\]

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## ✔ Equation 2: \( x^2 - 15x + 3 = -51 \)

Move all terms to left:
\[
x^2 - 15x + 3 + 51 = 0 \Rightarrow x^2 - 15x + 54 = 0
\]

Factor:
\[
(x - 6)(x - 9) = 0
\]
Solutions:
\[
x = 6, \quad x = 9
\]

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## ✔ Equation 3: \( x^2 + x - 70 = 2 \)

Move all terms to left:
\[
x^2 + x - 72 = 0
\]

Factor:
\[
(x + 9)(x - 8) = 0
\]
Solutions:
\[
x = -9, \quad x = 8
\]

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## ✔ Equation 4: \( x^2 + 5x - 10 = 14 \)

Move all terms to left:
\[
x^2 + 5x - 24 = 0
\]

Factor:
\[
(x + 8)(x - 3) = 0
\]
Solutions:
\[
x = -8, \quad x = 3
\]

---

## ✔ Equation 5: \( x^2 - 8x - 4 = 5 \)

Move all terms to left:
\[
x^2 - 8x - 9 = 0
\]

Factor:
\[
(x - 9)(x + 1) = 0
\]
Solutions:
\[
x = 9, \quad x = -1
\]

---

## ✔ Equation 6: \( x^2 - 2x - 35 = 28 \)

Move all terms to left:
\[
x^2 - 2x - 63 = 0
\]

Factor:
\[
(x - 9)(x + 7) = 0
\]
Solutions:
\[
x = 9, \quad x = -7
\]

---

## ✔ Equation 7: \( x^2 - 3x = -2 \)

Move all terms to left:
\[
x^2 - 3x + 2 = 0
\]

Factor:
\[
(x - 1)(x - 2) = 0
\]
Solutions:
\[
x = 1, \quad x = 2
\]

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## ✔ Equation 8: \( x^2 + 4x - 25 = 7 \)

Move all terms to left:
\[
x^2 + 4x - 32 = 0
\]

Factor:
\[
(x + 8)(x - 4) = 0
\]
Solutions:
\[
x = -8, \quad x = 4
\]

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## ✔ Equation 9: \( x^2 + 9x = -8 \)

Move all terms to left:
\[
x^2 + 9x + 8 = 0
\]

Factor:
\[
(x + 1)(x + 8) = 0
\]
Solutions:
\[
x = -1, \quad x = -8
\]

---

## ✔ Equation 10: \( x^2 - 6 = 43 \)

Move all terms to left:
\[
x^2 - 49 = 0
\]

Factor (difference of squares):
\[
(x - 7)(x + 7) = 0
\]
Solutions:
\[
x = 7, \quad x = -7
\]

---

## ✔ Equation 11: \( x^2 - 4x - 1 = 4 \)

Move all terms to left:
\[
x^2 - 4x - 5 = 0
\]

Factor:
\[
(x - 5)(x + 1) = 0
\]
Solutions:
\[
x = 5, \quad x = -1
\]

---

## ✔ Equation 12: \( x^2 + 6x - 15 = 12 \)

Move all terms to left:
\[
x^2 + 6x - 27 = 0
\]

Factor:
\[
(x + 9)(x - 3) = 0
\]
Solutions:
\[
x = -9, \quad x = 3
\]

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## 📝 Final Answer Summary:

Here are the solutions for each equation:

1. \( x^2 + 4x + 3 = 0 \) → x = -1, -3

2. \( x^2 - 15x + 3 = -51 \) → x = 6, 9

3. \( x^2 + x - 70 = 2 \) → x = -9, 8

4. \( x^2 + 5x - 10 = 14 \) → x = -8, 3

5. \( x^2 - 8x - 4 = 5 \) → x = 9, -1

6. \( x^2 - 2x - 35 = 28 \) → x = 9, -7

7. \( x^2 - 3x = -2 \) → x = 1, 2

8. \( x^2 + 4x - 25 = 7 \) → x = -8, 4

9. \( x^2 + 9x = -8 \) → x = -1, -8

10. \( x^2 - 6 = 43 \) → x = 7, -7

11. \( x^2 - 4x - 1 = 4 \) → x = 5, -1

12. \( x^2 + 6x - 15 = 12 \) → x = -9, 3

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✔ All equations have been solved using factoring after rearranging into standard form. Each has two real solutions (as expected for quadratics), and all factor nicely — no need for the quadratic formula here!

Let me know if you’d like to see any step explained further!