MaxWong

13x + 3 = -10

13x = -13

x = -1.

See your original post.

$(\sqrt2 + i\sqrt3)\times i\sqrt6 - \sqrt2\times(2i\sqrt2 - 3) + i\times (\dfrac{6}{\sqrt6}-\dfrac{6}{\sqrt3})\\ = x - y + z$

I am going to divide it into 3 parts.

x = (sqrt2 + i sqrt3) * i sqrt6 

y = sqrt2 *(2i sqrt2 - 3)

z = i * (6/sqrt6 - 6/sqrt3)

x = $(\sqrt2 + i\sqrt3) \times i\sqrt6 = i\sqrt{12} + i^2\sqrt{18}=2i\sqrt3 - 3\sqrt2$

y = $\sqrt2 \times (2i\sqrt2 - 3)=4i - 3\sqrt2$

z = $i\times (\dfrac{6}{\sqrt6}-\dfrac{6}{\sqrt3})=i\times (\dfrac{6-6\sqrt2}{\sqrt6})=i\times (\sqrt{6}-\sqrt{12})=i\sqrt6-2i\sqrt3$

      x - y + z

 $=(2i\sqrt3 - 2\sqrt2) -(4i-3\sqrt2)+(i\sqrt6 -2i\sqrt3)\\ =2i\sqrt3 - 2\sqrt2 - 4i + 3\sqrt2 + i\sqrt6 - 2i\sqrt3\\ = \sqrt2 +i(\sqrt6-4)$

Therefore the original equation is sqrt2 + i(sqrt6 - 4)

$x^2=169\\ x^2 + 0 x - 169 = 0\\ x = \dfrac{-0\pm\sqrt{0^2-4(1)(-169)}}{2(1)}\\ \;\;=\dfrac{\pm 26}{2}\\ \;\;=13 \text{ OR }-13$

Thanks for the answer, but my level requires cross-method, but I never understand it :(

I don't really understand cross-method, I am confused when I am doing it, it made me cross, but not made me doing factorization :(

There are infinitely many numbers between 2.71 and 2.72

2.71000000000000001, 2.71000000000000002, 2.71000000000000003...............