# How do you find the roots, real and imaginary, of y= 8x^2 - 10x + 14-(3x-1)^2  using the quadratic formula?

Mar 29, 2017

$x = - 2 + \sqrt{17} , - 2 - \sqrt{17}$

#### Explanation:

$y = 8 {x}^{2} - 10 x + 14 - {\left(3 x - 1\right)}^{2}$

$y = 8 {x}^{2} - 10 x + 14 - \left(9 {x}^{2} - 6 x + 1\right)$

$y = 8 {x}^{2} - 9 {x}^{2} - 10 x + 6 x + 14 - 1$

$y = - {x}^{2} - 4 x + 6 x + 13$

to find it root when $y = 0$
$0 = - {x}^{2} - 4 x + 6 x + 13$
$a = - 1 , b = - 4 \mathmr{and} c = 13$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(- 1\right) \left(13\right)}}{2 \left(- 1\right)}$

$x = \frac{\left(4\right) \pm \sqrt{\left(16 + 52\right)}}{- 2}$

x = (4 +- sqrt(68))/(-2) = (4 +- sqrt(4 * 17))/(-2) = (4 +- 2 sqrt( 17))/(-2

x = -[(2 +- sqrt(17)]

$x = - 2 + \sqrt{17} , - 2 - \sqrt{17}$