How do you solve #x^2+5x-4=0# using the quadratic formula?

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Answer 1 · 2016-03-30T16:33:46

#x= color(blue)( (-5+sqrt(41))/2#

#x= color(blue)( (-5-sqrt(41))/2#

#x^2 + 5x - 4 = 0#

#a=1, b=5, c=-4#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (5)^2-(4*1 * (-4))#

# = 25 + 16 = 41 #

The solutions are found using the formula:

#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-5)+-sqrt(41))/(2*1) = (-5+-sqrt(41))/2#

#x= color(blue)( (-5+sqrt(41))/2#

#x= color(blue)( (-5-sqrt(41))/2#