Answer
$x=-\frac{7}{4}\pm\frac{\sqrt{17}}{4}$
Work Step by Step
We complete the square:
$2x^{2}+7x+4=0$
$x^{2}+\frac{7}{2}x+2=0$
$x^{2}+\frac{7}{2}x=-2$
$x^{2}+\frac{7}{2}x+\frac{49}{16}=-2+\frac{49}{16}=\frac{17}{16}$
$(x+\frac{7}{4})^{2}=\frac{17}{16}$
$x+\displaystyle \frac{7}{4}=\pm\sqrt{\frac{17}{16}}=\pm\frac{\sqrt{17}}{4}$
$x=-\frac{7}{4}\pm\frac{\sqrt{17}}{4}$
