Quadratic Equation

For the quadratic equation in the question, $a=1$ , $b=2$ and $c=3$ . As $2^2 - 4(3)(1)$ is negative, and negative numbers don't have real square roots (notice that the quadratic formula requires $\sqrt{b^2-4ac}$ ), there are hence $\boxed{0}$ solutions.

The discriminant is negative so there will be non-real roots.

$b^2 = 4$ , which is smaller than $4ac$ , because $4ac = 4*1*3 = 12$ . So if we are to use the formula $\sqrt{b^2 - 4ac}$ , the number will be imaginary.