To solve the quadratic equation \(2x^2 + 7x - 15 = 0\) by factoring, we can first look for two numbers that multiply to \(2 \cdot (-15) = -30\) (the product of the coefficient of \(x^2\) and the constant term) and add up to \(7\) (the coefficient of \(x\)).
The numbers that satisfy these conditions are \(10\) and \(-3\) since:
- \(10 \cdot (-3) = -30\)
- \(10 + (-3) = 7\)
Now, we can rewrite the middle term \(7x\) using these two numbers:
\[ 2x^2 + 10x - 3x - 15 = 0 \]
Next, we can factor by grouping:
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Group the first two terms and the last two terms: \[ (2x^2 + 10x) + (-3x - 15) = 0 \]
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Factor out the common factors in each group: \[ 2x(x + 5) - 3(x + 5) = 0 \]
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Now, factor out the common binomial factor \((x + 5)\): \[ (2x - 3)(x + 5) = 0 \]
Now, we can set each factor equal to zero to find the solutions:
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For \(2x - 3 = 0\): \[ 2x = 3 \implies x = \frac{3}{2} \]
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For \(x + 5 = 0\): \[ x = -5 \]
Thus, the solutions to the quadratic equation \(2x^2 + 7x - 15 = 0\) are: \[ x = \frac{3}{2} \quad \text{and} \quad x = -5 \]