Answer
$x=\dfrac{1}{2}$
$x=-\dfrac{7}{2}$
Work Step by Step
$x^{2}+3x-\dfrac{7}{4}=0$
Let's take the independent term to the right side of the equation:
$x^{2}+3x=\dfrac{7}{4}$
Remember that, in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation, $b=3$:
$x^{2}+3x+(\dfrac{3}{2})^{2}=\dfrac{7}{4}+(\dfrac{3}{2})^{2}$
$x^{2}+3x+\dfrac{9}{4}=\dfrac{7}{4}+\dfrac{9}{4}$
$x^{2}+3x+\dfrac{9}{4}=\dfrac{16}{4}$
$x^{2}+3x+\dfrac{9}{4}=4$
We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes:
$(x+\dfrac{3}{2})^{2}=4$
Take the square root of both terms:
$\sqrt{(x+\dfrac{3}{2})^{2}}=\sqrt{4}$
$x+\dfrac{3}{2}=\pm2$
Solve for $x$:
$x=-\dfrac{3}{2}\pm2$
The two solutions are:
$x=-\dfrac{3}{2}+2=\dfrac{1}{2}$ and $x=-\dfrac{3}{2}-2=-\dfrac{7}{2}$
