How do you solve #x^2 + 3x - 5 = 0#?

1 Answer
Nov 1, 2015

Answer:

x = #(-3 + sqrt29)/2# or #x = (-3 - sqrt29)/2#

Explanation:

This equation does not easily factorise so using the general equation #y = ax^2 + bx +c# and the formula

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

then if # y = x^2 +3x-5#

a =1, b = 3 and c = -5

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

which comes to the given answer.
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