How do you solve the quadratic with complex numbers given $2x^2-6x+7=0$ ?

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Answer 1 · 2016-09-30T11:17:45

$x=3/2+isqrt5/2$ or $x=3/2-isqrt5/2$

$2x^2-6x+7=0$ - we first ty to make complete square as in $(x+a)^2$

$2x^2-6x+7=0$

$hArrx^2-3x+7/2=0$

or $x^2-2xx3/2xx x+(3/2)^2-(3/2)^2+7/2=0$

or $(x-3/2)^2-9/4+14/4=0$

or $(x-3/2)^2+5/4=0$

or $(x-3/2)^2-(sqrt5/2xxi)^2=0$

As this is of the form $a^2-b^2=(a+b)(a-b)$

The above is equal to

$(x-3/2-isqrt5/2)(x-3/2+isqrt5/2)0$

Hence $x-3/2-isqrt5/2=0$ or $x-3/2+isqrt5/2=0$

i.e. $x=3/2+isqrt5/2$ or $x=3/2-isqrt5/2$

How do you solve the quadratic with complex numbers given 2x^2-6x+7=02x^2-6x+7=0?