How do you solve #7x ^ { 2} - 56= 42x + 6#?

1 Answer
Nghi N
Jul 23, 2017

#x = 4 +- (5sqrt35)/7#

Explanation:

Use the quadratic formula ni graphic form (Socratic, Google search).
#y = 7x^2 - 42x - 62 = 0#
#D = d^2 = b^2 - 4ac = 1764 + 1736 = 3500# --> #d = +- 10sqrt35#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 56/14 +- 10sqrt35/14 = 4 +- (5sqrt35)/7#

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