Questions: MATH-1314-198 (1) Objective: solve quadratic equations using the quadratic formula. Level 0 Let's review a concept needed to answer your homework question. Solve the equation using the quadratic formula. 4x^2 - x - 14 = 0 The solution(s) to the quadratic equation ax^2 + bx + c = 0, a ≠ 0 are given by the quadratic formula x = (-b ± sqrt(b^2 - 4ac))/(2a) Since the equation is in standard form, identify the values of a, b, and c in the equation. What is the value of a? a = 4 What is the value of b? b = ☐

To solve the quadratic equation \(4x^2 - x - 14 = 0\) using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\) from the equation. Here, \(a = 4\), \(b = -1\), and \(c = -14\). We then substitute these values into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions for \(x\).

Dado el ecuación cuadrática \(4x^2 - x - 14 = 0\), identificamos los coeficientes:

• \(a = 4\)
• \(b = -1\)
• \(c = -14\)

El discriminante se calcula usando la fórmula \(b^2 - 4ac\): \[ \text{discriminante} = (-1)^2 - 4 \cdot 4 \cdot (-14) = 1 + 224 = 225 \]

La fórmula cuadrática es: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Sustituyendo los valores de \(a\), \(b\), y \(c\): \[ x = \frac{-(-1) \pm \sqrt{225}}{2 \cdot 4} = \frac{1 \pm 15}{8} \]

Calculamos las dos soluciones: \[ x_1 = \frac{1 + 15}{8} = \frac{16}{8} = 2.0 \] \[ x_2 = \frac{1 - 15}{8} = \frac{-14}{8} = -1.75 \]

\[ \boxed{x_1 = 2.0} \] \[ \boxed{x_2 = -1.75} \]