How do you solve #4x² = 2 – 7x#?

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Answer 1 · 2016-04-24T07:32:03

The solutions are:
#x= (-7+sqrt(17))/8#
#x= (-7-sqrt(17))/8#

#4x^2 = 2 - 7x#

#4x^2 + 7x - 2= 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=4, b=7, c= - 2#

The Discriminant is given by:

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

# = (7)^2-(4* 4 * (-2))#

# = 49 - 32 = 17#

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

#x = ((-7)+-sqrt(17))/(2*4) = (-7+-sqrt(17))/8#

The solutions are:
#x= (-7+sqrt(17))/8#
#x= (-7-sqrt(17))/8#