How do you solve #2x^2 + x + 7 = 0#?

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Answer 1 · 2016-03-09T14:10:05

The solutions are:
# x=(-1+sqrt(-55))/4#

# x=(-1-sqrt(-55))/4#

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

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=2, b=1, c=7#

The Discriminant is given by:

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

# = (1)^2-(4*2*7)#

# = 1 - 56 = - 55#

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

#x = ((-1)+-sqrt(-55))/(2*2) = (-1+-sqrt(-55))/4#

The solutions are:
# x=(-1+sqrt(-55))/4#

# x=(-1-sqrt(-55))/4#