How do you solve #y = 16x^2 + 40x + 25# using the quadratic formula?

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Answer 1 · 2016-04-29T14:46:32

The solution is
#x = -5/4#

#y = 16x^2 + 40x +25#

#16x^2 + 40x +25 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=16, b=40, c=25#

The Discriminant is given by:

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

# = (40)^2-(4* 16 * 25)#

# = 1600 - 1600 = 0#

The solutions are normally found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-40)+-sqrt(0))/(2*16) = (-40+-0)/32#

#x = -40/32#

#x = -5/4#