How do you solve #x^2+x-10=0#?

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Answer 1 · 2016-02-26T16:17:17

The solutions are:
# x= (-1+sqrt(41))/2#

# x= (-1-sqrt(41))/2#

#x^2 +x -10=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=1, c=-10#

The Discriminant is given by:

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

# = (1)^2-(4* 1*-10)#

# = 1 +40 = 41#

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

#x = ((-1)+-sqrt(41))/(2*1) = (-1+-sqrt(41))/2#

The solutions are:
# x= (-1+sqrt(41))/2#

# x= (-1-sqrt(41))/2#