Respuesta :

apologiabiology
minus 12x both sides
0=5x^2-12x+9
use quadratic formula
for 0=ax^2+bx+c
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]

given
5x^2-12x+9
a=5
b=-12
c=9
remember: i=√-1

x=[tex] \frac{-(-12)+/- \sqrt{(-12)^2-4(5)(9)} }{2(5)} [/tex]
x=[tex] \frac{12+/- \sqrt{144-180} }{10} [/tex]
x=[tex] \frac{12+/- \sqrt{-36} }{10} [/tex]
x=[tex] \frac{12+/- (\sqrt{-1})(\sqrt{36}) }{10} [/tex]
x=[tex] \frac{12+/- i(\sqrt{36}) }{10} [/tex]
x=[tex] \frac{12+/- 6i }{10} [/tex]
x=[tex] \frac{6+/- 3i }{5} [/tex]

the roots are
x=[tex] \frac{6+ 3i }{5} [/tex] and [tex] \frac{6- 3i }{5} [/tex]