## College Algebra (6th Edition)

$x = 2 \frac{+}{} 2\sqrt {5}$
$$(x - 2)^{2} = 20$$ $(x^{2} - 4x + 4) = 20$ $x^{2} - 4x + 4 - 20 = 0$ $x^{2} - 4x - 16 = 0$ Since this equation is not easily factorized, we can use the quadratic formula: ($\frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$) $x = \frac{-(-4) \frac{+}{} \sqrt {(-4)^{2} - 4(1)(-16)}}{2(1)} = \frac{4 \frac{+}{} \sqrt {(16 +64)}}{2} = \frac{4 \frac{+}{} \sqrt {80}}{2} = \frac{4 \frac{+}{} 4\sqrt {5}}{2} = \frac{4(1 \frac{+}{} \sqrt {5})}{2} = 2 \frac{+}{} 2\sqrt {5}$