How do you evaluate #y= - 2x ^ { 2} - 2x + 3; if y = \frac { 7} { 2}#?

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Answer 1 · 2018-06-17T20:30:28

#x=-1/2#

#7/2 = -2x^2 -2x + 3#

#2x^2 +2x-3+3,5 = 0#

#2x^2 +2x+1/2 = 0#

#x=(-2+-(2^2-4.2.1/2))/2.2#

#x=(-2+-0)/4#

#x=-2/4#

#x=-1/2#