Solve $\color{blue}{x^2-3x-3 = 0}$ using the Quadratic Formula.
Step 1: Read the values of $ a $, $ b $, and $ c $ from the quadratic equation: $ a $ is the number in front of $ x^2 $, $ b $ is the number in front of $ x $, $ c $ is the number at the end. In our case:
$$ a = 1, \,\, b = -3, \,\, c = -3 $$Step 2: Plug in the values for $ a $, $ b $, and $ c $ into the quadratic formula.
$$ \begin{aligned} x_1,x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\[1 em] x_1,x_2 &= \frac{ -(-3) \pm \sqrt{ (-3)^2 - 4 \cdot 1 \cdot (-3)} }{ 2 \cdot 1 } \end{aligned} $$Step 3: Simplify expression under the square root.
$$ x_1,x_2 = \frac{ 3 \pm \sqrt{ 21 } }{ 2 } $$Step 4: Solve for $ x $
$$ \begin{aligned} & \color{blue}{ x_1 = \frac{ 3~-~\sqrt{ 21 } }{ 2 } = \frac{ 3 }{ 2 }-\frac{\sqrt{ 21 }}{ 2 } } \\\\ & \color{blue}{ x_2 = \frac{ 3~+~\sqrt{ 21 } }{ 2 } = \frac{ 3 }{ 2 }+\frac{\sqrt{ 21 }}{ 2 } } \end{aligned} $$