Respuesta :

apologiabiology
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]

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