Answer: [tex]\bold{b)\quad \dfrac{-2\pm \sqrt{14}}{2}}[/tex]
Step-by-step explanation:
[tex]2x^2+4x-5=0\qquad \implies\qquad a=2,\ b=4,\ c=-5\\\\\text{The quadratic equation is not factorable so use the Quadratic Formula:}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-4\pm \sqrt{4^2-4(2)(-5)}}{2(2)}\\\\\\.\ =\dfrac{-4\pm \sqrt{16+40}}{4}\\\\\\.\ =\dfrac{-4\pm \sqrt{56}}{4}\\\\\\.\ =\dfrac{-4\pm 2\sqrt{14}}{4}\\\\\\.\ =\large\boxed{\dfrac{-2\pm \sqrt{14}}{2}}[/tex]
