How do you solve the equation: #x^2 + 24x + 90 = 0#?

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Answer 1 · 2015-07-03T06:35:43

Check the discriminant, then use the quadratic formula to find:

#x = -12 +-3sqrt(6)#.

That is:

#x = -12-3sqrt(6)# or #x = -12+3sqrt(6)#

#f(x) = x^2+24x+90# is of the form

#ax^2+bx+c# with #a=1#, #b=24# and #c=90#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 24^2-(4xx1xx90) = 576 - 360#

#= 216 = 6^2*6#

Since #Delta# is positive, but not a perfect square, #f(x) = 0# has two distinct irrational real solutions, given by the quadratic formula:

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

#= (-b+-sqrt(Delta))/(2a)#

#= (-24+-sqrt(216))/2#

#= (-24+-6sqrt(6))/2#

#=-12+-3sqrt(6)#