## College Algebra (10th Edition)

The solution set is {$\frac{-1-\sqrt 5}{4},\frac{-1+\sqrt 5}{4}$}.
$4x^2+2x-1=0$ $a=4, b=2, c=-1$ $D=b^2-4ac=(2)^2-4\cdot4(-1)=4+16=20>0$ The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-2\pm\sqrt 20}{2\cdot4}$ Use the quadratic formula. $x=\frac{-2\pm2\sqrt 5}{8}$ $x=\frac{-1+\sqrt 5}{4}$,$x=\frac{-1-\sqrt 5}{4}$ The solution set is {$\frac{-1-\sqrt 5}{4},\frac{-1+\sqrt 5}{4}$}.