To solve the equation \( 2a^2 + 4a - 6 = -3a \), we first rearrange the terms to one side of the equation:
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Add \( 3a \) to both sides: \[ 2a^2 + 4a + 3a - 6 = 0 \] This simplifies to: \[ 2a^2 + 7a - 6 = 0 \]
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Now, we want to solve the quadratic equation \( 2a^2 + 7a - 6 = 0 \). We can use the quadratic formula, which states that for an equation in the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = 7 \), and \( c = -6 \).
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Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot (-6) = 49 + 48 = 97 \]
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Now substitute \( a \), \( b \), and the discriminant into the quadratic formula: \[ a = \frac{-7 \pm \sqrt{97}}{2 \cdot 2} = \frac{-7 \pm \sqrt{97}}{4} \]
Therefore, the solutions for \( a \) are: \[ a = \frac{-7 + \sqrt{97}}{4} \quad \text{and} \quad a = \frac{-7 - \sqrt{97}}{4} \]
These are the two real solutions in simplest form.