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

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Answer 1 · 2017-05-31T02:27:44

#x# Can either #= -4.56155 or -0.43845#

#ax^2 + bx + c = 0#

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

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

#a = 1#
#b = 5#
#c = -2#

#x = (-5+-sqrt(5^2 + 4 xx 1 xx -2))/(2 xx 1#

#x = (-5+-sqrt(25 + -8))/(2#

#x = (-5+-sqrt17)/(2#

#x_1 = (-5 - 4.1231)/(2#

#x_1 = (-9.1231)/(2#

#color(blue)(x_1 = -4.56155#

#x_2 = (-5 + 4.1231)/(2#

#x_2 = -0.8769/(2#

#color(blue)(x_2 = -0.43845#