## College Algebra (10th Edition)

$\displaystyle x=\frac{-1\pm\sqrt{5}}{2}$
We are given: $x^{2}+x=1$ $x^{2}+x-1=0$ We solve using the quadratic formula ($a=1,\ b=1,\ c=-1$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\displaystyle x=\frac{-1\pm\sqrt{(1)^{2}-4*1*-1}}{2*1}$ $\displaystyle x=\frac{-1\pm\sqrt{1+4}}{2}$ $\displaystyle x=\frac{-1\pm\sqrt{5}}{2}$