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

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

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Answer 1 · 2016-11-03T15:12:21

ans. $x = \frac{1 + i \sqrt{3}}{2}$ $x=(1+isqrt3)/2$ or $\frac{1 - i \sqrt{3}}{2}$ $(1-isqrt3)/2$ .

Explanation:

$2 x^{2} - 2 x + 7 = 5$ $2x^2-2x+7=5$
or, $2 x^{2} - 2 x + 2 = 0$ $2x^2-2x+2=0$
$or, x^{2} - x + 1 = 0$ $or,x^2-x+1=0$ $(i)$ $(i)$
now according to the rule if $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ then $x = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+sqrt(b^2-4ac))/(2a)$ or $\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b-sqrt(b^2-4ac))/(2a)$
hence from the equation $(i)$ $(i)$ we get the ans. $x = \frac{1 + i \sqrt{3}}{2}$ $x=(1+isqrt3)/2$ or $\frac{1 - i \sqrt{3}}{2}$ $(1-isqrt3)/2$ .
here $i = \sqrt{- 1}$ $i=sqrt(-1)$ .

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

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