To solve the quadratic equation \(12b^2 + 21b + 10 = 3b^2\), we first rearrange it into standard form \(ax^2 + bx + c = 0\).
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Move \(3b^2\) to the left side of the equation: \[ 12b^2 - 3b^2 + 21b + 10 = 0 \] Simplifying this, we get: \[ 9b^2 + 21b + 10 = 0 \]
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The quadratic equation is now in the form \(9b^2 + 21b + 10 = 0\), where \(a = 9\), \(b = 21\), and \(c = 10\).
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We can apply the quadratic formula, which is given by: \[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] Here \(A = 9\), \(B = 21\), and \(C = 10\).
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First, calculate the discriminant \(B^2 - 4AC\): \[ B^2 = 21^2 = 441 \] \[ 4AC = 4 \cdot 9 \cdot 10 = 360 \] So: \[ B^2 - 4AC = 441 - 360 = 81 \]
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Since the discriminant is positive, we have two real solutions. Now, substituting in the quadratic formula: \[ b = \frac{-21 \pm \sqrt{81}}{2 \cdot 9} \]
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Calculate \(\sqrt{81}\): \[ \sqrt{81} = 9 \]
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Substitute \(\sqrt{81}\) back into the formula: \[ b = \frac{-21 \pm 9}{18} \]
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Now we calculate the two possible solutions:
- For the positive case: \[ b = \frac{-21 + 9}{18} = \frac{-12}{18} = -\frac{2}{3} \]
- For the negative case: \[ b = \frac{-21 - 9}{18} = \frac{-30}{18} = -\frac{5}{3} \]
Thus, the solutions to the quadratic equation \(9b^2 + 21b + 10 = 0\) are: \[ b = -\frac{2}{3} \quad \text{and} \quad b = -\frac{5}{3} \]