X^2+6x-18<0 помогите плизз

$x^{2} +6x-18\ \textless \ 0\\ x^{2} +6x-18=0\\D=36+72=108 \\ \sqrt{D} = \sqrt{36*3} =6 \sqrt{3} \\ \\ x_{1} = \frac{-6+6 \sqrt{3} }{2} =-3+3 \sqrt{3} \\ \\ x_{2} = \frac{-6-6 \sqrt{3} }{2} =-3-3 \sqrt{3} \\ \\ (x+3+ \sqrt{3} )(x+3-3 \sqrt{3} )\ \textless \ 0$
+ - +
------------|------------------------|-------------------------------> x
-3-3√3 -3+√3

x ∈ $(-3-3 \sqrt{3} ;-3+3 \sqrt{3} )$