How do you solve #2x ^ { 2} + x = 5#?

1 Answer
Nghi N
Dec 26, 2016

-0.53, 1.35

Explanation:

Bring to standard form:
#y = 2x^2 + x - 5 = 0#
#D = d^2 = 1 + 40 = 41 --> d = +- sqrt41#
There are 2 real rots:
#x = -b/(2a) +- d/(2a) = - 1/4 +- sqrt41/4#

#x1 = (- 1 + 6.40)4 = 5.40/4 = 1.35#
#x2 = (- 1 - 6.40)/14 = -7.40/14 = - 0.53#

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