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

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Answer 1 · 2018-01-04T14:51:42

no real solutions

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

$:.y=2x^2-3x+4+4(x^2+2x+1)$

$:.y=2x^2-3x+4+4x^2+8x+4$

$:.y=4x^2+5x+8$

$:.y=(-b+-sqrt(b^2-4ac))/(2a)$

$:.y=(-(5)+-sqrt((5)^2-4(6))(8))/(2(6))$

$:.y=(-5+-sqrt(-167))/(12)$

no real solutions.

How do you find the roots, real and imaginary, of y= 2x^2 - 3x + 4+ 4(-x-1)^2 y=2x2−3x+4+4(−x−1)2y= 2x^2 - 3x + 4+ 4(-x-1)^2 using the quadratic formula?