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

1 Answer
May 5, 2016

Answer:

#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.