The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -1.8003\\[1 em]x_3 &= -2.6032\\[1 em]x_4 &= 0.7017+1.0006i\\[1 em]x_5 &= 0.7017-1.0006i \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x^2 }$ from $ x^6+3x^5+7x^2 $ and solve two separate equations:
$$ \begin{aligned} x^6+3x^5+7x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( x^4+3x^3+7 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ x^4+3x^3+7 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ x^4+3x^3+7 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.