Here are the solutions to each question from the image:

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1. Solve the equation $x^2 - 11x + 28 = 0$.

This is a quadratic equation in standard form: $ax^2 + bx + c = 0$.
We solve it using factoring:

$x^2 - 11x + 28 = 0$

Find two numbers that multiply to $28$ and add to $-11$. These numbers are $-4$ and $-7$.
$x^2 - 11x + 28 = (x - 4)(x - 7) = 0$

Set each factor equal to 0:
$x - 4 = 0 \quad \text{or} \quad x - 7 = 0$ $x = 4 \quad \text{or} \quad x = 7$

Solution: $x = 4, 7$.

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2. Solve $2x^2 + 4x = 0$.

Factor out the greatest common factor:
$2x^2 + 4x = 0 \implies 2x(x + 2) = 0$

Set each factor equal to 0:
$2x = 0 \quad \text{or} \quad x + 2 = 0$ $x = 0 \quad \text{or} \quad x = -2$

Solution: $x = 0, -2$.

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3. Find the exact solution of $x^2 - 14x = -40$.

Rearrange into standard quadratic form:
$x^2 - 14x + 40 = 0$

Factor the quadratic: Find two numbers that multiply to $40$ and add to $-14$. These numbers are $-10$ and $-4$.
$x^2 - 14x + 40 = (x - 10)(x - 4) = 0$

Set each factor equal to 0:
$x - 10 = 0 \quad \text{or} \quad x - 4 = 0$ $x = 10 \quad \text{or} \quad x = 4$

Solution: $x = 10, 4$.

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4. Find the exact solution of $-x^2 + 3x + 7 = 0$.

First, multiply through by $-1$ to simplify:
$x^2 - 3x - 7 = 0$

This cannot be factored easily, so use the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1, b = -3, c = -7$.

Substitute into the formula:
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-7)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 + 28}}{2}$ $x = \frac{3 \pm \sqrt{37}}{2}$

Solution: $x = \frac{3 + \sqrt{37}}{2}$ and $x = \frac{3 - \sqrt{37}}{2}$.

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5. Simplify $\sqrt{25x^{20}y^{14}}$.

Split the square root:
$\sqrt{25x^{20}y^{14}} = \sqrt{25} \cdot \sqrt{x^{20}} \cdot \sqrt{y^{14}}$

Simplify each part:
$\sqrt{25} = 5, \quad \sqrt{x^{20}} = x^{10}, \quad \sqrt{y^{14}} = y^7$

Combine:
$\sqrt{25x^{20}y^{14}} = 5x^{10}y^7$

Solution: $5x^{10}y^7$.

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6. Simplify $\sqrt{36x^4y^{20}}$.

Split the square root:
$\sqrt{36x^4y^{20}} = \sqrt{36} \cdot \sqrt{x^4} \cdot \sqrt{y^{20}}$

Simplify each part:
$\sqrt{36} = 6, \quad \sqrt{x^4} = x^2, \quad \sqrt{y^{20}} = y^{10}$

Combine:
$\sqrt{36x^4y^{20}} = 6x^2y^{10}$

Solution: $6x^2y^{10}$.

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Do you want any specific question explained in further detail? Let me know!

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5 Related Questions

1. Solve $x^2 - 9x + 20 = 0$ using factoring.
2. Solve $3x^2 + 6x = 0$ by factoring out the GCF.
3. Simplify $\sqrt{49x^{16}y^{12}}$.
4. Use the quadratic formula to solve $2x^2 - 4x - 6 = 0$.
5. Solve $x^2 + 6x + 9 = 0$.

Tip:

Always check if a quadratic can be factored first; if not, use the quadratic formula!