Question Number 32396 by saru53424@gmail.com last updated on 24/Mar/18
$${roots} \\ $$$$\mathrm{2}{x}×\boldsymbol{{x}}+\boldsymbol{{x}}+\mathrm{3} \\ $$
Answered by $@ty@m last updated on 24/Mar/18
$$\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{3}=\mathrm{0} \\ $$$${a}=\mathrm{2}\:{b}=\mathrm{1}\:{c}=\mathrm{3} \\ $$$${D}={b}^{\mathrm{2}} −\mathrm{4}{ac}=−\mathrm{23} \\ $$$${x}=\frac{−{b}\pm\sqrt{{D}}}{\mathrm{2}{a}} \\ $$$$=\frac{−\mathrm{1}\pm\sqrt{−\mathrm{23}}}{\mathrm{4}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}{i} \\ $$