Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.