How do you find the roots, real and imaginary, of y=-14 x^2 +18x +16-(x-3)^2 using the quadratic formula?

1 Answer
sente
Aug 6, 2016

x in {4/5-sqrt(249)/15, 4/5+sqrt(249)/15}

Explanation:

The quadratic formula gives us the solutions to a quadratic equation of the form ax^2+bx+c=0. Thus, to use it, we must work the equation into that form.

-14x^2+18x+16-(x-3)^2

= -14x^2+18x+16-(x^2-6x+9)

= -14x^2+18x+16-x^2+6x-9

=-15x^2+24x+7

Thus, we can now look for the roots as the solutions to ax^2+bx+c=0, with a=-15, b=24, c=7

Plugging our values into the formula, we get

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

=(-24+-sqrt(24^2-4(-15)(7)))/(2(-15))

=-(-24+-sqrt(576+420))/30

=(24+-sqrt(996))/30

=4/5+-sqrt(249)/15

Thus, the roots of the given equation are {4/5-sqrt(249)/15, 4/5+sqrt(249)/15}

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