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

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Answer 1 · 2015-12-18T03:02:46

$x=(1+-sqrt73)/12$

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

$y=ax^2+bx+c$

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 $2x-1$ term.

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

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

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

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

Thus,

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

so,

$x=(-1+-sqrt(1^2-4(-6)(3)))/(2(-6))$

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

$x=(1+-sqrt73)/12$

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