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

Jun 4, 2016

1.435 and .232

#### Explanation:

first you want to simplify the equation by expanding the brackets and then collecting like terms.

$- {x}^{2} - x + {\left(2 x - 1\right)}^{2}$

$- {x}^{2} - x + \left(2 x - 1\right) \left(2 x - 1\right)$

$- {x}^{2} - x + \left(4 {x}^{2} - 2 x - 2 x + 1\right)$

$- {x}^{2} - x + 4 {x}^{2} - 4 x + 1$

$3 {x}^{2} - 5 x + 1$

$\therefore a = 3 , b = - 5 \mathmr{and} c = 1$

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

$\frac{- \left(- 5\right) \pm \sqrt{- {5}^{2} - 4 \left(3\right) \left(1\right)}}{2 \left(3\right)}$

$\frac{5 \pm \sqrt{25 - 12}}{6}$

$\frac{5 \pm \sqrt{13}}{6}$

$\frac{5 + 3.61}{6}$ and $\frac{5 - 3.61}{6}$

$1.435$ and $.232$