Question

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

(If there is more than one solution, separate them with commas.)

Answer 1

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

so

for
ax^2+bx+c=0

`x=(-b+/-√(b^2-4ac))/(2a)`
so
given
2x^2-3x+9=0
a=2
b=-3
c=9


`x=(-(-3)+/-√((-3)^2-4(2)(9)))/(2(2))`

`x=(3+/-√(9-72))/(4)`

`x=(3+/-√(-63))/(4)`

`x=(3+/-(√(-1))(√(63)))/(4)`

`x=(3+/-(i)(3√(7)))/(4)`

`x=(3+/-3i√(7))/(4)`

the 2 complex solutions (in form a+bi) are

`x=(3)/(4)+(3i√(7))/(3)` and
`x=(3)/(4)-(3i√(7))/(3)`