Expand and rearrange $(x+4)(x-3)=2x+8$, then apply the quadratic formula. The roots are:

Expanding: $x^2+x-12=2x+8$, so $x^2-x-20=0.x=\frac{1\pm \sqrt{1+80}}{2}=\frac{1\pm 9}{2}=5$ or −4. Check: $(5+4)(5-3)=18=2(5)+8$, and $(-4+4)(-4-3)=0=2(-4)+8$. Setting each bracket to 0 — which gives 4 and −3 — is only allowed when the product itself is 0, and here it equals $2x+8$.