How do you solve 15x^2+7x-55 using the quadratic formula?

1 Answer
George C.
May 5, 2016

x=(-7+-sqrt(3349))/30

Explanation:

Strictly speaking, I guess you would like to solve 15x^2+7x-55=0 or in other words find the zeros of 15x^2+7x-55. You do not "solve" a quadratic expression.

That having been said, the equation 15x^2+7x-55 = 0 is of the form ax^2+bx+c = 0 with a=15, b=7 and c=55.

This has roots (solutions) given by the quadratic formula:

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

=(-7+-sqrt((-7)^2-(4*15*(-55))))/(2*15)

=(-7+-sqrt(49+3300))/30

=(-7+-sqrt(3349))/30

The square root sqrt(3349) does not simplify further since the prime factorisation of 3349 is 17*197, which contains no square factors.

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