How do you solve 2x ^ { 2} - 2x + 7= 5?

1 Answer
Nov 3, 2016

ans. x=(1+isqrt3)/2 or (1-isqrt3)/2.

Explanation:

2x^2-2x+7=5
or,2x^2-2x+2=0
or,x^2-x+1=0 (i)
now according to the rule if ax^2+bx+c=0 then x=(-b+sqrt(b^2-4ac))/(2a) or (-b-sqrt(b^2-4ac))/(2a)
hence from the equation (i) we get the ans.x=(1+isqrt3)/2 or (1-isqrt3)/2.
here i=sqrt(-1).