# How do you find the roots, real and imaginary, of y= 2x^2 - 2x -1  using the quadratic formula?

May 31, 2016

$x = \frac{1 + \sqrt{3}}{2}$ and $x = \frac{1 - \sqrt{3}}{2}$

#### Explanation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where a, b, and c correspond to the coefficients in the equation

$y = a {x}^{2} + b x + c$

In this case, $a = 2 , b = - 2 ,$ and $c = - 1$. Plugging these values into the formula gives

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 2 \cdot \left(- 1\right)}}{2 \cdot 2} = \frac{2 \pm \sqrt{12}}{4}$

Now because the discriminant (the thing in the square root) does not equal zero, there are two answers, namely

$x = \frac{2 + \sqrt{12}}{4}$ and $x = \frac{2 - \sqrt{12}}{4}$

which can be simplified to

$x = \frac{1 + \sqrt{3}}{2}$ and $x = \frac{1 - \sqrt{3}}{2}$