How do you find the roots, real and imaginary, of $y= (3x+14)x+30x-x^2+14$ using the quadratic formula?

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Answer 1 · 2018-05-26T06:31:18

See explanation.

To calculate the roots we first have to transform the function into $y=ax^2+bx+c$ .

$y=(3x+14)x+30x-x^2+14$

$y=3x^2+14x+30x-x^2+14$

$y=2x^2+30x+14$

Now we can use the quadratic formula:

First calculate the determinant:

$Delta=30^2-4*2*14=900-122=788$

The determinant is positive, so the function has two different real roots:

$x_1=(-b-sqrt(Delta))/(2a)=(-30-sqrt(788))/4$

$x_1=(-b+sqrt(Delta))/(2a)=(-30+sqrt(788))/4$

How do you find the roots, real and imaginary, of y= (3x+14)x+30x-x^2+14 y= (3x+14)x+30x-x^2+14 using the quadratic formula?