Answer :
[tex]2x^2+7x-3=0 \\ \\
a=2 \\ b=7 \\ c=-3 \\ b^2-4ac=7^2-4 \times 2 \times (-3) =49+24=73 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-7 \pm \sqrt{73}}{2 \times 2}=\frac{-7 \pm \sqrt{73}}{4} \\ \\
\boxed{x=\frac{-7+\sqrt{73}}{4} \hbox{ or } x=\frac{-7-\sqrt{73}}{4}}[/tex]
[tex]2x^2+7x-3=0\\\\
We\ are\ calculating\ delta\ to\ find\ out\ how\many\ real\ zeros \ are\ in \equation\\\\\ \Delta=b^2-4ac\\\\
a=2,\ b=7,\ c=-3 \\\\
\sqrt{\Delta}=7^2-4*2*(-3)=49+24=73\\\\
\sqrt{\Delta}=\sqrt{73}\\\\
x_1=\frac{-b-\sqrt{\Delta}}{2a}\ \ x_1=\frac{-7-\sqrt{73}}{4}
\\\\ or \\
x_2=\frac{-b+\sqrt{\Delta}}{2a}\ \ x_2=\frac{-7+\sqrt{73}}{4}
[/tex]
