How do you find the roots, real and imaginary, of $y=12x^2 -7x +28-(x-2)^2$ using the quadratic formula?

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Answer 1 · 2018-04-13T15:46:05

$x=3/22+-sqrt1047/22i$

The quadratic formula states that, for a quadratic in the form

$y=color(red)(a)x^2+color(green)(b)x+color(blue)(c) => x=(-color(green)(b)+-sqrt(color(green)(b)^2-4color(red)(a)color(blue)(c)))/(2color(red)(a))$

We need to put our given quadratic in the form $y=color(red)(a)x^2+color(green)(b)x+color(blue)(c)$

$y=12x^2-7x+28-(x-2)^2$

$=12x^2-7x+28-(x^2-4x+4)$

$=color(red)(11)x^2color(green)(-3)x+color(blue)(24)$

From the quadratic formula:

$x=(-color(green)((-3))+-sqrt(color(green)((-3))^2-4xxcolor(red)(11)xxcolor(blue)(24)))/(2xxcolor(red)(11))$

$=(3+-sqrt(-1047))/22$

We can separate $sqrt(-1047)$ into $sqrt1047xxsqrt(-1)$ . Since $i=sqrt(-1)$ :

$x=(3+-sqrt1047 i)/22$

$x=3/22+-sqrt1047/22i$

How do you find the roots, real and imaginary, of y=12x^2 -7x +28-(x-2)^2 y=12x^2 -7x +28-(x-2)^2 using the quadratic formula?