Question

1. solve (24x^2 + 23x + 28 = 0).

(a) (b) (c)
(x = \frac{-b pm sqrt{b^2 - 4ac}}{2a})
(x = \frac{-23 pm sqrt{23^2 - 4(24)(28)}}{2(24)})

Explanation:

Step1: Identify coefficients

For quadratic equation \(ax^2 + bx + c = 0\), here \(a = 24\), \(b = 23\), \(c = 28\).

Step2: Calculate discriminant \(\Delta\)

\(\Delta = b^2 - 4ac = 23^2 - 4\times24\times28\)
\(= 529 - 2688 = -2159\)

Step3: Apply quadratic formula

Quadratic formula: \(x=\frac{-b\pm\sqrt{\Delta}}{2a}\)
Substitute values: \(x=\frac{-23\pm\sqrt{-2159}}{2\times24}=\frac{-23\pm i\sqrt{2159}}{48}\) (since \(\sqrt{-1}=i\))

Answer:

The solutions are \(x = \frac{-23 + i\sqrt{2159}}{48}\) and \(x = \frac{-23 - i\sqrt{2159}}{48}\)