Grade 10 - Worksheets - Mathematics - Free Printable

Let's solve each of the quadratic equations step by step. The goal is to factorize and solve for $ x $ in each equation.

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1. $ -2x^2 + 9x + 20 = -6 $

Step 1: Bring all terms to one side:
$$
-2x^2 + 9x + 20 + 6 = 0 \Rightarrow -2x^2 + 9x + 26 = 0
$$

Step 2: Multiply both sides by $-1$ to make leading coefficient positive:
$$
2x^2 - 9x - 26 = 0
$$

Step 3: Factor:
We need two numbers that multiply to $ 2 \times (-26) = -52 $ and add to $-9$.
Try: $-13$ and $4$: $-13 \times 4 = -52$, $-13 + 4 = -9$ → Yes!

Split middle term:
$$
2x^2 - 13x + 4x - 26 = 0
\Rightarrow x(2x - 13) + 2(2x - 13) = 0
\Rightarrow (x + 2)(2x - 13) = 0
$$

Solutions:
$$
x + 2 = 0 \Rightarrow x = -2 \\
2x - 13 = 0 \Rightarrow x = \frac{13}{2}
$$

✔ Answer: $ x = -2 $ or $ x = \frac{13}{2} $

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2. $ -x^2 - 2x + 7 = -17 $

Step 1: Move all terms to one side:
$$
-x^2 - 2x + 7 + 17 = 0 \Rightarrow -x^2 - 2x + 24 = 0
$$

Step 2: Multiply by $-1$:
$$
x^2 + 2x - 24 = 0
$$

Step 3: Factor:
Find two numbers that multiply to $-24$ and add to $2$: $6$ and $-4$

$$
x^2 + 6x - 4x - 24 = 0 \Rightarrow x(x + 6) - 4(x + 6) = 0 \Rightarrow (x - 4)(x + 6) = 0
$$

Solutions:
$$
x = 4 \quad \text{or} \quad x = -6
$$

✔ Answer: $ x = 4 $ or $ x = -6 $

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3. $ 2x^2 - 3x - 4 = 5 $

Step 1: Move all terms:
$$
2x^2 - 3x - 4 - 5 = 0 \Rightarrow 2x^2 - 3x - 9 = 0
$$

Step 2: Factor:
Multiply $2 \times (-9) = -18$, find two numbers that multiply to $-18$ and add to $-3$: $-6$ and $3$

Split:
$$
2x^2 - 6x + 3x - 9 = 0 \Rightarrow 2x(x - 3) + 3(x - 3) = 0 \Rightarrow (2x + 3)(x - 3) = 0
$$

Solutions:
$$
2x + 3 = 0 \Rightarrow x = -\frac{3}{2} \\
x - 3 = 0 \Rightarrow x = 3
$$

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

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4. $ 4x^2 - 20x + 16 = -8 $

Step 1: Move all terms:
$$
4x^2 - 20x + 16 + 8 = 0 \Rightarrow 4x^2 - 20x + 24 = 0
$$

Step 2: Divide all terms by 4:
$$
x^2 - 5x + 6 = 0
$$

Step 3: Factor:
$$
x^2 - 5x + 6 = (x - 2)(x - 3) = 0
$$

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

✔ Answer: $ x = 2 $ or $ x = 3 $

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5. $ 2x^2 + 22x + 13 = -23 $

Step 1: Move all terms:
$$
2x^2 + 22x + 13 + 23 = 0 \Rightarrow 2x^2 + 22x + 36 = 0
$$

Step 2: Divide by 2:
$$
x^2 + 11x + 18 = 0
$$

Step 3: Factor:
Find two numbers that multiply to $18$ and add to $11$: $9$ and $2$

$$
(x + 9)(x + 2) = 0
$$

Solutions:
$$
x = -9 \quad \text{or} \quad x = -2
$$

✔ Answer: $ x = -9 $ or $ x = -2 $

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6. $ x^2 - x - 44 = 12 $

Step 1: Move all terms:
$$
x^2 - x - 44 - 12 = 0 \Rightarrow x^2 - x - 56 = 0
$$

Step 2: Factor:
Find two numbers that multiply to $-56$ and add to $-1$: $-8$ and $7$

$$
x^2 - 8x + 7x - 56 = 0 \Rightarrow x(x - 8) + 7(x - 8) = 0 \Rightarrow (x + 7)(x - 8) = 0
$$

Solutions:
$$
x = -7 \quad \text{or} \quad x = 8
$$

✔ Answer: $ x = -7 $ or $ x = 8 $

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7. $ 2x^2 - 21x + 28 = 12 $

Step 1: Move all terms:
$$
2x^2 - 21x + 28 - 12 = 0 \Rightarrow 2x^2 - 21x + 16 = 0
$$

Step 2: Factor:
Multiply $2 \times 16 = 32$, find two numbers that multiply to $32$ and add to $-21$: $-16$ and $-5$

Split:
$$
2x^2 - 16x - 5x + 16 = 0 \Rightarrow 2x(x - 8) -1(x - 8) = 0 \Rightarrow (2x - 1)(x - 8) = 0
$$

Solutions:
$$
2x - 1 = 0 \Rightarrow x = \frac{1}{2} \\
x - 8 = 0 \Rightarrow x = 8
$$

✔ Answer: $ x = \frac{1}{2} $ or $ x = 8 $

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8. $ -2x^2 + 17x - 12 = -9 $

Step 1: Move all terms:
$$
-2x^2 + 17x - 12 + 9 = 0 \Rightarrow -2x^2 + 17x - 3 = 0
$$

Step 2: Multiply by $-1$:
$$
2x^2 - 17x + 3 = 0
$$

Step 3: Try factoring: $2x^2 - 17x + 3$

We need two numbers that multiply to $2 \times 3 = 6$ and add to $-17$. No such integers exist.

Use quadratic formula:
$$
x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(2)(3)}}{2(2)} = \frac{17 \pm \sqrt{289 - 24}}{4} = \frac{17 \pm \sqrt{265}}{4}
$$

Since $\sqrt{265}$ is irrational, no rational factorization.

✔ Answer: $ x = \frac{17 \pm \sqrt{265}}{4} $ (No rational solutions)

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9. $ -x^2 - 2x + 19 = -29 $

Step 1: Move all terms:
$$
-x^2 - 2x + 19 + 29 = 0 \Rightarrow -x^2 - 2x + 48 = 0
$$

Step 2: Multiply by $-1$:
$$
x^2 + 2x - 48 = 0
$$

Step 3: Factor:
Find two numbers that multiply to $-48$ and add to $2$: $8$ and $-6$

$$
x^2 + 8x - 6x - 48 = 0 \Rightarrow x(x + 8) -6(x + 8) = 0 \Rightarrow (x - 6)(x + 8) = 0
$$

Solutions:
$$
x = 6 \quad \text{or} \quad x = -8
$$

✔ Answer: $ x = 6 $ or $ x = -8 $

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10. $ x^2 + 10x + 4 = -5 $

Step 1: Move all terms:
$$
x^2 + 10x + 4 + 5 = 0 \Rightarrow x^2 + 10x + 9 = 0
$$

Step 2: Factor:
$$
x^2 + 10x + 9 = (x + 1)(x + 9) = 0
$$

Solutions:
$$
x = -1 \quad \text{or} \quad x = -9
$$

✔ Answer: $ x = -1 $ or $ x = -9 $

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## ✔ Final Answers:

| Equation | Solution |
|--------|---------|
| 1. $ -2x^2 + 9x + 20 = -6 $ | $ x = -2 $ or $ x = \frac{13}{2} $ |
| 2. $ -x^2 - 2x + 7 = -17 $ | $ x = 4 $ or $ x = -6 $ |
| 3. $ 2x^2 - 3x - 4 = 5 $ | $ x = -\frac{3}{2} $ or $ x = 3 $ |
| 4. $ 4x^2 - 20x + 16 = -8 $ | $ x = 2 $ or $ x = 3 $ |
| 5. $ 2x^2 + 22x + 13 = -23 $ | $ x = -9 $ or $ x = -2 $ |
| 6. $ x^2 - x - 44 = 12 $ | $ x = -7 $ or $ x = 8 $ |
| 7. $ 2x^2 - 21x + 28 = 12 $ | $ x = \frac{1}{2} $ or $ x = 8 $ |
| 8. $ -2x^2 + 17x - 12 = -9 $ | $ x = \frac{17 \pm \sqrt{265}}{4} $ |
| 9. $ -x^2 - 2x + 19 = -29 $ | $ x = 6 $ or $ x = -8 $ |
| 10. $ x^2 + 10x + 4 = -5 $ | $ x = -1 $ or $ x = -9 $ |

> ⚠️ Note: Problem 8 does not factor nicely over integers. Use the quadratic formula if exact values are needed.

Let me know if you'd like a visual explanation or steps drawn out!