Solve for \( x \) using the quadratic formula. \( \begin{array}{r}2 x^{2}+14 x+12=0 \\ x=[?],\end{array} \) Quadratic Formula: \( x=\frac{-b \pm \sqrt{b^{2}}-4 a c}{2 a} \)

To solve the quadratic equation \(2x^2 + 14x + 12 = 0\) using the quadratic formula, we identify \(a = 2\), \(b = 14\), and \(c = 12\). Plugging these values into the formula, we first calculate the discriminant: \[ b^2 - 4ac = 14^2 - 4(2)(12) = 196 - 96 = 100. \] Now substituting back, we get: \[ x = \frac{-14 \pm \sqrt{100}}{2 \cdot 2} = \frac{-14 \pm 10}{4}. \] This gives us two solutions: \[ x_1 = \frac{-4}{4} = -1, \] \[ x_2 = \frac{-24}{4} = -6. \] Thus, the solutions are \( x = -1 \) and \( x = -6 \). When using the quadratic formula, remember to accurately calculate the discriminant, as it determines the nature of the roots. A positive discriminant indicates two real and distinct solutions, a zero discriminant indicates one real solution, and a negative discriminant leads to complex solutions. Always double-check your arithmetic to avoid simple calculation errors!