How do you find the roots, real and imaginary, of $y= (3/4x+1460)x+300$ using the quadratic formula?

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Answer 1 · 2018-03-15T12:14:28

$x=(-2920+-20sqrt(21307))/3$

We take $(3/4x+1460)x+300$ and expand it:

$3/4x^2+1460x+300$

We equate this to $0$ , and use the quadratic formula to solve it.

The formula states that:

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

Here, $a=3/4,b=1460,c=300$ . Inputting:

$x=(-1460+-sqrt(1460^2-4(3/4)(300)))/(2*3/4)$

$x=(-1460+-sqrt(2131600-900))/(3/2)$

$x=(-1460+-sqrt(2130700))/(3/2)$

$x=(-2920+-20sqrt(21307))/3$

How do you find the roots, real and imaginary, of y= (3/4x+1460)x+300 y= (3/4x+1460)x+300 using the quadratic formula?