## College Algebra 7th Edition

$x^{2}-3x+9=0$ We solve using the quadratic formula ($a=1, b=-3, c=9$): $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{--3\pm\sqrt{(-3)^{2}-4*1*9}}{2*1}=\frac{3\pm\sqrt{9-36}}{2}=\frac{3\pm\sqrt{-27}}{2}$ Since we have the square root of a negative, there are no real solutions.