How do you solve #35x^ + 98x + 40=0#?

1 Answer
Nov 11, 2015

Answer:

Assuming the intended equation was #35x^2+98x+40=0#, use the quadratic formula to find:

#x=-7/5+-sqrt(1001)/35#

Explanation:

#35x^2+98x+40# is of the form #ax^2+bx+c# with #a=35#, #b=98# and #c=40#.

This has zeros given by the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)=(-98+-sqrt(98^2-(4*35*40)))/(2*35)=-7/5+-sqrt(49^2-1400)/35#

#=-7/5+-sqrt(1001)/35#