College Algebra (6th Edition)

$$0 = x^2 + 6x + 8$$ $$becomes$$ $$x=\frac{-(6)\frac{+}{} \sqrt{(6)^2-4(1)(8)}}{2(1)}$$
The quadratic formula $x=\frac{-b\frac{+}{} \sqrt{b^2-4ac}}{2a}$ merely requires direct substitution of the coefficients of any quadratic equation in the form $0 = ax^2 + bx + c$. For the case of $0 = x^2 + 6x + 8$: $$x=\frac{-(6)\frac{+}{} \sqrt{(6)^2-4(1)(8)}}{2(1)}$$ $$x=\frac{-6\frac{+}{} \sqrt{36-32}}{2}$$ $$x=\frac{-6\frac{+}{} \sqrt{4}}{2}$$ $$x=\frac{-6\frac{+}{} 2}{2}$$ $$x = -4$$ $$or$$ $$x = -2$$