Sure, let's solve each of these equations step-by-step.
### 1. Equation 1: [tex]\( 3x - 2x^2 = 7 \)[/tex]
First, we rearrange the equation to the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ -2x^2 + 3x - 7 = 0 \][/tex]
With [tex]\( a = -2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -7 \)[/tex].
To solve this quadratic equation, we use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
However, the solutions are:
[tex]\[ x_1 = \frac{3}{4} - \frac{\sqrt{47}i}{4}, \][/tex]
[tex]\[ x_2 = \frac{3}{4} + \frac{\sqrt{47}i}{4}. \][/tex]
### 2. Equation 2: [tex]\( (x + 3)(x + 4) = 0 \)[/tex]
To solve this equation, we set each factor equal to zero:
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
The solutions are:
[tex]\[ x_1 = -3, \][/tex]
[tex]\[ x_2 = -4. \][/tex]
### 3. Equation 3: [tex]\( 2x(x - 3) = 15 \)[/tex]
First, we expand the left side and move all terms to one side to get it into a standard quadratic form:
[tex]\[ 2x^2 - 6x = 15 \][/tex]
[tex]\[ 2x^2 - 6x - 15 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ x_1 = \frac{3}{2} - \frac{\sqrt{39}}{2}, \][/tex]
[tex]\[ x_2 = \frac{3}{2} + \frac{\sqrt{39}}{2}. \][/tex]
### 4. Equation 4: [tex]\( (x + 7)(x - 7) = -3x \)[/tex]
First, we expand the left side and move all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 49 = -3x \][/tex]
[tex]\[ x^2 + 3x - 49 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ x_1 = -\frac{3}{2} + \frac{\sqrt{205}}{2}, \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{205}}{2} - \frac{3}{2}. \][/tex]
### 5. Equation 5: [tex]\( 2x(x + 4) = (x - 3)(x - 3) \)[/tex]
First, we expand both sides and then move all terms to one side to form a standard quadratic equation:
[tex]\[ 2x^2 + 8x = x^2 - 6x + 9 \][/tex]
[tex]\[ 2x^2 + 8x - x^2 + 6x - 9 = 0 \][/tex]
[tex]\[ x^2 + 14x - 9 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ x_1 = -7 + \sqrt{58}, \][/tex]
[tex]\[ x_2 = -\sqrt{58} - 7. \][/tex]
These are the solutions for each of the equations given.
