$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2B+\cos^2C-1=\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
$$x^2+2x\cos B\cos C+\cos^2B+\cos^2C-1=0$$
$$\iff x^2+x\{\cos(B+C)+\cos(B-C)\}+\cos(B+C)\cos(B-C)=0$$
Now use $y^2+(a+b)y+ab=(y+a)(y+b)$ and $\cos(B+C)=\cos(\pi-A)=-\cos A$
