#### Answer

The solutions are $x=1$ and $x=-\dfrac{5}{3}$

#### Work Step by Step

$3x^{2}+2x=5$
Take out common factor $3$ from the left side:
$3\Big(x^{2}+\dfrac{2}{3}x\Big)=5$
Take $3$ to divide the right side:
$x^{2}+\dfrac{2}{3}x=\dfrac{5}{3}$
Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=\dfrac{2}{3}$
$x^{2}+\dfrac{2}{3}x+\Big(\dfrac{2}{2\cdot3}\Big)^{2}=\dfrac{5}{3}+\Big(\dfrac{2}{2\cdot3}\Big)^{2}$
$x^{2}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{5}{3}+\dfrac{1}{9}$
$x^{2}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{16}{9}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x+\dfrac{1}{3}\Big)^{2}=\dfrac{16}{9}$
Take the square root of both sides:
$\sqrt{\Big(x+\dfrac{1}{3}\Big)^{2}}=\pm\sqrt{\dfrac{16}{9}}$
$x+\dfrac{1}{3}=\pm\dfrac{4}{3}$
Solve for $x$:
$x=-\dfrac{1}{3}\pm\dfrac{4}{3}$
The two solutions are:
$x=-\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{3}{3}=1$
$x=-\dfrac{1}{3}-\dfrac{4}{3}=-\dfrac{5}{3}$