Answer
Two solutions, complex conjugates (not real)
Work Step by Step
For $ax^{2}+bx+c=0$, we can find solutions using the
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
The radicand in the formula is called the discriminant.
$D=b^{2}-4ac$
D is positive $\Rightarrow$ there are two distinct real solutions.
D is zero $\Rightarrow$ there is one (double) real solution.
D is negative $\Rightarrow$ two solutions, complex conjugates (not real).
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Here,
$ 4x^{2}-2x+3=0\quad\rightarrow \left\{\begin{array}{l}
a=4\\
b=-2\\
c=3
\end{array}\right.$
$D=b^{2}-4ac=(-2)^{2}-4(4)(3)=4-48=-44$
$D$ is negative $\Rightarrow$ two solutions, complex conjugates (not real)
