Question

Find all complex solutions of 2x^2-3x+9=0.

Answer 1

We use quadratic formula for that one
remember that √-1=i

so

for
ax^2+bx+c=0
$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$
so
given
2x^2-3x+9=0
a=2
b=-3
c=9

$x=\frac{-(-3)+/-\sqrt{(-3)^2-4(2)(9)}}{2(2)}$
$x=\frac{3+/-\sqrt{9-72}}{4}$
$x=\frac{3+/-\sqrt{-63}}{4}$
$x=\frac{3+/-(\sqrt{-1})(\sqrt{63})}{4}$
$x=\frac{3+/-(i)(3\sqrt{7})}{4}$
$x=\frac{3+/-3i\sqrt{7}}{4}$

the 2 complex solutions (in form a+bi)  are
$x=\frac{3}{4}+\frac{3i\sqrt{7}}{3}$ and $x=\frac{3}{4}-\frac{3i\sqrt{7}}{3}$