Question
Solve the following quadratic equation:
$6x^2 + x - 12 = 0$
$6x^2 + x - 12 = 0$
✓
Answer
$6x^2 + x - 12 = 0$
$\Rightarrow 6x^2 + 9x - 8x - 12 = 0$
$\Rightarrow 3x(2x + 3) - 4(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x - 4) = 0$
$\Rightarrow 2x + 3 = 0$ or $3x - 4 = 0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{4}{3}$
Hence, $\frac{-3}{2}$ and $\frac{4}{3}$ are the roots of $6x^2 + x - 12 = 0$
$\Rightarrow 6x^2 + 9x - 8x - 12 = 0$
$\Rightarrow 3x(2x + 3) - 4(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x - 4) = 0$
$\Rightarrow 2x + 3 = 0$ or $3x - 4 = 0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{4}{3}$
Hence, $\frac{-3}{2}$ and $\frac{4}{3}$ are the roots of $6x^2 + x - 12 = 0$
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