[tex]\displaystyle\it\\\text{We have the equation: }\\\\ax^2+bx+c=0\\\\\text{We find the roots:}\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_{2}=\frac{-b-\sqrt{b^2-4ac}}{2a}\\\\\text{If } b^2-4ac>0 \text{ then }x_1, x_2\in~R\\\\\text{If } b^2-4ac=0 \text{ then }x_1=x_2\in~R\\\\\text{If } b^2-4ac<0 \text{ then }x_1, x_2\in~C[/tex]
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