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Solve for x using quaratic formula:2x^2+7x-3=0?

Sagot :

naǫ
[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]
luana
[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]
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