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

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Answer 1 · 2017-03-29T15:53:17

$x = -2 + sqrt 17, -2 - sqrt17$

$y = 8 x^2 - 10 x + 14 - (3 x -1)^2$

$y = 8 x^2 - 10 x + 14 - (9x^2 -6x +1)$

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

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

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

$x = (-b +- sqrt(b^2 -4ac))/(2a)$
$x = (-(-4) +- sqrt((-4)^2 -4(-1)(13)))/(2(-1))$

$x = ((4) +- sqrt((16 + 52)))/(-2)$

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

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

$x = -2 + sqrt 17, -2 - sqrt17$

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