$$(x+1)(x-2)(x-8)(x+5)+360 = 0$$
Answer
$$ \begin{matrix}x_1 = 4 & x_2 = 7.34662 & x_3 = -3.67331+1.21641i \\[1 em] x_4 = -3.67331-1.21641i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x+1)(x-2)(x-8)(x+5)+360 &= 0&& \text{simplify left side} \\[1 em](x^2-2x+x-2)(x-8)(x+5)+360 &= 0&& \\[1 em](x^2-x-2)(x-8)(x+5)+360 &= 0&& \\[1 em](x^3-8x^2-x^2+8x-2x+16)(x+5)+360 &= 0&& \\[1 em](x^3-9x^2+6x+16)(x+5)+360 &= 0&& \\[1 em]x^4-4x^3-39x^2+46x+80+360 &= 0&& \\[1 em]x^4-4x^3-39x^2+46x+440 &= 0&& \\[1 em] \end{aligned} $$
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