#### Answer

$x=(\frac{-3+\sqrt 17}{2}, \frac{-3-\sqrt 17}{2})$

#### Work Step by Step

Step-1 : Add $2$ from both sides of the equation
$x^2+3x=2$
Step -2 : Add the square of half of the co-efficient of $x$ to both sides.
Co-efficient of $x =3$
Half of $3 = \frac{1}{2}×3=\frac{3}{2}$
Square of $\frac{3}{2}$ is $\frac{3}{2}×\frac{3}{2}=\frac{9}{4}$
The equation becomes $x^2+3x+\frac{9}{4}=2+\frac{9}{4}$
Step-3 Factor the trinomial and simplify the right hand side.
$(x+\frac{3}{2})^2=\frac{8+9}{4}$
$(x+\frac{3}{2})^2=\frac{17}{4}$
Step-4 Use the square root property and solve for $x$
$(x+\frac{3}{2})=±\sqrt\frac{17}{4} $
Step-5 Subtract \frac{3}{2} on both the sides
$x=-\frac{3}{2}±\sqrt \frac{17}{4}$
Step-6 Simplify the right hand side
$x=-\frac{3}{2}±\frac{\sqrt 17}{2}$
$x=\frac{-3±\sqrt 17}{2}$
Therefore the solution set is $(\frac{-3+\sqrt 17}{2}, \frac{-3-\sqrt 17}{2})$