Answer
The solutions are $x=-4$ and $x=\dfrac{1}{2}$
Work Step by Step
$2x^{2}+7x=4$
Take out common factor $2$ from the left side of the equation:
$2\Big(x^{2}+\dfrac{7}{2}x\Big)=4$
Take $2$ to divide the right side:
$x^{2}+\dfrac{7}{2}x=\dfrac{4}{2}$
$x^{2}+\dfrac{7}{2}x=2$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=\dfrac{7}{2}$:
$x^{2}+\dfrac{7}{2}x+\Big(\dfrac{7}{2\cdot2}\Big)^{2}=2+\Big(\dfrac{7}{2\cdot2}\Big)^{2}$
$x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}=2+\dfrac{49}{16}$
$x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}=\dfrac{81}{16}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{16}$
Take the square root of both sides:
$\sqrt{\Big(x+\dfrac{7}{4}\Big)^{2}}=\pm\sqrt{\dfrac{81}{16}}$
$x+\dfrac{7}{4}=\pm\dfrac{9}{4}$
Solve for $x$:
$x=\pm\dfrac{9}{4}-\dfrac{7}{4}$
$x=\dfrac{9}{4}-\dfrac{7}{4}=\dfrac{2}{4}=\dfrac{1}{2}$
$x=-\dfrac{9}{4}-\dfrac{7}{4}=-\dfrac{16}{4}=-4$
The solutions are $x=-4$ and $x=\dfrac{1}{2}$