How do you solve #-x^2 - 3x + 5 = 0# using the quadratic formula?

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Answer 1 · 2016-02-27T07:33:06

#x=( -3-sqrt(29))/(2) or x=( -3+sqrt(29))/(2)#

Given equation= # -x^2-3x+5=0#

Comparing with #ax^2+bx+c=0#

We get , #a= -1, b=-3, c=5#

#x=( -b+-sqrt(b^2-4ac))/(2a)#

#x=( 3+-sqrt(-3^2-4xx-1xx5))/(2xx-1)#

#x=( 3+-sqrt(9+20))/(-2)#

#x=( 3+-sqrt(29))/(-2)#

#x=( 3+sqrt(29))/(-2) or x=( 3-sqrt(29))/(-2)#

#x=( -3-sqrt(29))/(2) or x=( -3+sqrt(29))/(2)#