How do you solve $2x²-(5/2)x+2=0$ ?

Question

1113 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T07:58:44

$x=5/8+isqrt39/8$ or $5/8-isqrt39/8$

$2x^2-5/2x+2=0$ can be written as

$2(x^2-5/4 x+1)=0$ - dividing by $2$

or $x^2-5/4 x+1=0$

or $x^2-2xx5/8xx x+(5/8)^2-(5/8)^2+1=0$

or $(x-5/8)^2-25/64+1=0$

or $(x-5/8)^2+39/64=0$

or $(x-5/8)^2-(-1xx39/64)=0$

or $(x-5/8)^2-(isqrt39/8)^2=0$

or $(x-5/8+isqrt39/8)(x-5/8-isqrt39/8)=0$

Hence $x=5/8+isqrt39/8$ or $5/8-isqrt39/8$

How do you solve 2x²-(5/2)x+2=02x²-(5/2)x+2=0?