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The solutions are \( x=-\frac{2}{3} \) and \( x=-\frac{1}{2} \).
Solución
Solve the equation \( 6x^{2}+7x=-2 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(6x^{2}+7x=-2\)
- step1: Move the expression to the left side:
\(6x^{2}+7x+2=0\)
- step2: Factor the expression:
\(\left(2x+1\right)\left(3x+2\right)=0\)
- step3: Separate into possible cases:
\(\begin{align}&2x+1=0\\&3x+2=0\end{align}\)
- step4: Solve the equation:
\(\begin{align}&x=-\frac{1}{2}\\&x=-\frac{2}{3}\end{align}\)
- step5: Rewrite:
\(x_{1}=-\frac{2}{3},x_{2}=-\frac{1}{2}\)
The solutions to the equation \(6x^{2}+7x=-2\) are \(x=-\frac{2}{3}\) and \(x=-\frac{1}{2}\).
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Simplifique esta solución Extra Insights
To solve the equation \( 6x^2 + 7x + 2 = 0 \), we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = 7 \), and \( c = 2 \). 1. First, calculate the discriminant: \[ b^2 - 4ac = 7^2 - 4 \cdot 6 \cdot 2 = 49 - 48 = 1 \] 2. Now, plug the values into the quadratic formula: \[ x = \frac{-7 \pm \sqrt{1}}{2 \cdot 6} = \frac{-7 \pm 1}{12} \] 3. This gives us two possible solutions: \[ x_1 = \frac{-7 + 1}{12} = \frac{-6}{12} = -\frac{1}{2} \] \[ x_2 = \frac{-7 - 1}{12} = \frac{-8}{12} = -\frac{2}{3} \] So the solutions for \( x \) are \( x = -\frac{1}{2} \) and \( x = -\frac{2}{3} \).
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