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

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Answer 1 · 2015-08-12T05:58:48

The solutions are:
#color(blue)(x= (1+sqrt(-39))/4, x=(1-sqrt(-39))/4#

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

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=2, b=-1, c=5#

The Discriminant is given by:
#Delta=b^2-4*a*c#

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

# = 1-40=-39#

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

#x = (-(-1)+-sqrt(-39))/(2*2) = (1+-sqrt(-39))/4#

The solutions are:
#color(blue)(x= (1+sqrt(-39))/4, x=(1-sqrt(-39))/4#