# What are the zero(s) of: x^2 = 6x + 6 = 0?

Nov 9, 2015

$3 + \sqrt{15} , 3 - \sqrt{15}$

#### Explanation:

We can use the quadratic formula to find the zeros. We are given:

${x}^{2} = 6 x + 6$

We can arrange this into a quadratic equation:

${x}^{2} - 6 x - 6 = 0$

$x = \frac{- b \left(\frac{+}{-}\right) \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 1 , b = - 6 , c = - 6$
$x = \frac{- \left(- 6\right) \left(\frac{+}{-}\right) \sqrt{{\left(- 6\right)}^{2} - 4 \left(1\right) \left(- 6\right)}}{2 \left(1\right)} = \frac{6 \left(\frac{+}{-}\right) \sqrt{36 + 24}}{2}$
$x = \frac{6 \left(\frac{+}{-}\right) \sqrt{60}}{2} = \frac{6 \left(\frac{+}{-}\right) 2 \sqrt{15}}{2} = 3 \left(\frac{+}{-}\right) \sqrt{15}$