Question

$b^2 - 4ac$ is negative $
ightarrow$ 2 imaginary solutions
$x^2 + 3x + 5 = 0$

Explanation:

Step1: Identify coefficients

For $x^2 + 3x + 5 = 0$, $a=1$, $b=3$, $c=5$.

Step2: Compute discriminant

Calculate $b^2-4ac$:
$3^2 - 4\times1\times5 = 9 - 20$

Step3: Evaluate discriminant

$9 - 20 = -11$

Step4: Determine solution type

Since $b^2-4ac=-11<0$, there are 2 imaginary solutions.

Step5: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$x=\frac{-3\pm\sqrt{-11}}{2\times1}=\frac{-3\pm i\sqrt{11}}{2}$

Answer:

The equation has two imaginary solutions: $x=\frac{-3 + i\sqrt{11}}{2}$ and $x=\frac{-3 - i\sqrt{11}}{2}$