Questions: What is the solution to the following equation? x^2+3x+4=0 A. x=3 ; x=-1 B. x=1 ; x=-3 C. x=(3 ± √25)/2 D. x=(-3 ± √-7)/2

Solution Steps

The given equation is a quadratic equation of the form:

\[ x^2 + 3x + 4 = 0 \]

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\[ \Delta = b^2 - 4ac \]

For the equation \(x^2 + 3x + 4 = 0\), we have \(a = 1\), \(b = 3\), and \(c = 4\). Thus, the discriminant is:

\[ \Delta = 3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]

Since the discriminant \(\Delta = -7\) is negative, the quadratic equation has two complex solutions.

The solutions to the quadratic equation are given by the quadratic formula:

\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]

Substituting the values of \(a\), \(b\), and \(\Delta\):

\[ x = \frac{-3 \pm \sqrt{-7}}{2 \cdot 1} = \frac{-3 \pm \sqrt{-7}}{2} \]

Final Answer