How do you solve #8x^2+ 42=55x#?

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Answer 1 · 2016-03-23T16:52:33

The solutions are:
#x = color(blue)( 6 #

#x = color(blue)( 7/8 #

#8x^2+42=55x#

#8x^2 - 55x + 42=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=8, b=-55, c=42#

The Discriminant is given by:

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

# = (-55)^2-(4* 8 * 42)#

# = 3025 - 1344 = 1681#

The solutions are found using the formula:

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

#x = (-(-55)+-sqrt(1681))/(2*8) = (55+-41)/16#

#x = (55+41)/16 = 96/16 = color(blue)( 6 #

#x = (55 - 41)/16 = 14/16 = color(blue)( 7/8 #