Question

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

Answer 1

`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 \\\\ √(\Delta)=7^2-4*2*(-3)=49+24=73\\\\ √(\Delta)=√(73)\\\\ x_1=(-b-√(\Delta))/(2a)\ \ x_1=(-7-√(73))/(4) \\\\ or \\ x_2=(-b+√(\Delta))/(2a)\ \ x_2=(-7+√(73))/(4)`

Answer 2

`2x^2+7x-3=0 \\ \\ a=2 \\ b=7 \\ c=-3 \\ b^2-4ac=7^2-4 * 2 * (-3) =49+24=73 \\ \\ x=(-b \pm √(b^2-4ac))/(2a)=(-7 \pm √(73))/(2 * 2)=(-7 \pm √(73))/(4) \\ \\ \boxed{x=(-7+√(73))/(4) \hbox{ or } x=(-7-√(73))/(4)}`