we use quadratic formula for that one
remember that √-1=i
so
for
ax^2+bx+c=0
[tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]
so
given
2x^2-3x+9=0
a=2
b=-3
c=9
[tex]x=\frac{-(-3)+/-\sqrt{(-3)^2-4(2)(9)}}{2(2)}[/tex]
[tex]x=\frac{3+/-\sqrt{9-72}}{4}[/tex]
[tex]x=\frac{3+/-\sqrt{-63}}{4}[/tex]
[tex]x=\frac{3+/-(\sqrt{-1})(\sqrt{63})}{4}[/tex]
[tex]x=\frac{3+/-(i)(3\sqrt{7})}{4}[/tex]
[tex]x=\frac{3+/-3i\sqrt{7}}{4}[/tex]
the 2 complex solutions (in form a+bi) are
[tex]x=\frac{3}{4}+\frac{3i\sqrt{7}}{3}[/tex] and [tex]x=\frac{3}{4}-\frac{3i\sqrt{7}}{3}[/tex]
