Solve $(x-1) (x-2) (x+3) (x+4) +6=0$.

Given:  $(x-1) (x-2) (x+3) (x+4) +6=0$.

To do: To solve the given equation.

$(x−1)(x−2)(x+3)(x+4)+6=0$ 

$\Rightarrow (x−1)(x+3)(x−2)(x+4)+6=0$

$\Rightarrow (x^2+2x−3)(x^2+2x−8)+6=0$

Let $y=x^2+2x$

$( x^2+2x−3)( x^2+2x−8)+6=0$

$\Rightarrow( y−3)( y−8)+6=0$

$\Rightarrow y^2−11y+24+6=0$

$\Rightarrow y^2−11y+30=0$

$\Rightarrow ( y−6)( y−5)=0$

$\Rightarrow y=6$ or $y=5$

If $y=6$

$\Rightarrow x^2+2x=6$

$\Rightarrow x^2+2x−6=0$

$\Rightarrow x=\frac{−2\pm \sqrt{4+24}}{2}=\frac{−2\pm2\sqrt{7}}{2}$

$\Rightarrow x=−1\pm\sqrt{7}$

If $y=5$

$\Rightarrow x^2+2x=5$

$\Rightarrow x^2+2x−5=0$

$\Rightarrow x=\frac{−2 \pm \sqrt{4+20}}{2}=\frac{−2\pm2\sqrt{6}}{2}$

$\Rightarrow x=−1\pm\sqrt{6}$

$\therefore x=−1\pm\sqrt{7},\ x=−1\pm\sqrt{6}$ are the roots of the given equation.