# How do you find the roots, real and imaginary, of y=-(2x-1)^2 -2x^2 - 3x + 4  using the quadratic formula?

Dec 18, 2015

$x = \frac{1 \pm \sqrt{73}}{12}$

#### Explanation:

First, get the equation into the standard form of a quadratic equation

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

which can be very easily applied to the quadratic equation

x=(-b+-sqrt(b^2-4ac))/(2a

To get the equation into standard form, simplify by combining like terms.

First, square and expand the $2 x - 1$ term.

$y = - \left(2 x - 1\right) \left(2 x - 1\right) - 2 {x}^{2} - 3 x + 4$

$y = - \left(4 {x}^{2} - 4 x + 1\right) - 2 {x}^{2} - 3 x + 4$

$y = - 4 {x}^{2} + 4 x - 1 - 2 {x}^{2} - 3 x + 4$

$y = - 6 {x}^{2} + x + 3$

Thus,

$a = - 6$
$b = 1$
$c = 3$

so,

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(- 6\right) \left(3\right)}}{2 \left(- 6\right)}$

x=(-1+-sqrt(1+72))/(-12

$x = \frac{1 \pm \sqrt{73}}{12}$