Answer
$x=-\dfrac{1}{2}\pm\sqrt{\dfrac{23}{12}}$
Work Step by Step
$3x^{2}+3x=5$
Take out common factor $3$ from the left side of the equation:
$3(x^{2}+x)=5$
Take the $3$ to divide the right side of the equation:
$x^{2}+x=\dfrac{5}{3}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=1$
$x^{2}+x+\Big(\dfrac{1}{2}\Big)^{2}=\dfrac{5}{3}+\Big(\dfrac{1}{2}\Big)^{2}$
$x^{2}+x+\dfrac{1}{4}=\dfrac{5}{3}+\dfrac{1}{4}$
$x^{2}+x+\dfrac{1}{4}=\dfrac{23}{12}$
Factor the expression on the left side of the equation, which is a perfect square trinomial:
$\Big(x+\dfrac{1}{2}\Big)^{2}=\dfrac{23}{12}$
Take the square root of both sides of the equation:
$\sqrt{\Big(x+\dfrac{1}{2}\Big)^{2}}=\sqrt{\dfrac{23}{12}}$
$x+\dfrac{1}{2}=\pm\sqrt{\dfrac{23}{12}}$
Solve for $x$:
$x=-\dfrac{1}{2}\pm\sqrt{\dfrac{23}{12}}$
