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

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Answer 1 · 2017-05-02T07:42:40

#x=(-1)/(2)±(sqrt(15))/(2)#

#-2x^2+7=2x#

subtract #2x# from both sides and set the equation equal to #0#:
#-2x^2-2x+7=0#

plug the values #(a=-2, b=-2, c=7)# into the quadratic formula:

#x=(-(-2)±sqrt(-2^2-4(-2)(7)))/(2(-2))#

#x=(2±sqrt(60))/(-4)#

The discriminant #b2−4ac>0#
so, there are two real roots.

#x=(2±2sqrt(15))/(−4)#

#x=(2)/(-4)±(2sqrt(15))/(−4)#

#→x=(-1)/(2)±(sqrt(15))/(2)#