#### Answer

$x=\dfrac{1}{2}$ and $x=-4$

#### Work Step by Step

$2x^{2}+7x=4$
Take out common factor $2$ from the left side of the equation:
$2\Big(x^{2}+\dfrac{7}{2}x\Big)=4$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $2\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular case, $b=\dfrac{7}{2}$
$2\Big[x^{2}+\dfrac{7}{2}x+\Big(\dfrac{7}{4}\Big)^{2}\Big]=4+2\Big(\dfrac{7}{4}\Big)^{2}$
$2\Big(x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}\Big)=4+\dfrac{49}{8}$
$2\Big(x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}\Big)=\dfrac{81}{8}$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$2\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{8}$
Take the $2$ to divide the right side of the equation:
$\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{16}$
Take the square root of both sides of the equation:
$\sqrt{\Big(x+\dfrac{7}{4}\Big)^{2}}=\pm\sqrt{\dfrac{81}{16}}$
$x+\dfrac{7}{4}=\pm\dfrac{9}{4}$
Solve for $x$:
$x=\dfrac{-7\pm9}{4}$
The two solutions are:
$x=\dfrac{-7+9}{4}=\dfrac{1}{2}$
$x=\dfrac{-7-9}{4}=-4$