## Intermediate Algebra for College Students (7th Edition)

{$-4-\sqrt {21},-4+\sqrt {21}$}
Given: $x^2+8x-5=0$ This can be re-written as: $x^2+8x=5$ We need to add $(4)^2$ to complete the square. Thus, $x^2+8x+16=5+16$ or, $(x+4)^2=21$ or, $(x+4)=\pm \sqrt {21}$ or, $x=-4- \sqrt {21},-4+\sqrt {21}$ Hence, solution set is $x=${$-4-\sqrt {21},-4+\sqrt {21}$}