How do you find the roots, real and imaginary, of $y=4x^2 -7x -(x-2)^2$ using the quadratic formula?

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Answer 1 · 2016-04-13T06:30:21

$x=(3+ sqrt(57))/6$
$x=(3- sqrt(57))/6$

Given -

$y=4x^2-7x-(x-2)^2$

Simplify

$y=4x^2-7x-(x^2-4x+4)$
$y=4^2-7x-x^2+4x-4$
$y=3x^2-3x-4$

It is in a quadratic form -

Then If $sqrt(b^2-(4ac))$ is positive, its roots are real otherwise imaginary.

$sqrt((-3)^2-(4 xx 3 xx (-4))$
$sqrt(9-(-48)$
$sqrt(9+48)$
$sqrt(57)$

Since $sqrt(57)$ is positive, the roots are real

$x=(-(-3)+-sqrt(57))/(2 xx 3)$

$x=(3+ sqrt(57))/6$
$x=(3- sqrt(57))/6$

How do you find the roots, real and imaginary, of y=4x^2 -7x -(x-2)^2 y=4x^2 -7x -(x-2)^2 using the quadratic formula?