## Destinyyyy one year ago Simplify the equation. Then solve by the quadratic formula..

1. anonymous

3x^2-4x+3=0

2. anonymous

x= -(-4)+- square root (-4)^2 -4(3)(3)/2(3)

3. anonymous

4+square root 16-36/6 4+- square root -20/6 6(4+-4i square root 5)/6 cancel out the 6 My final answer is 4+-2i square root 5 but its wrong

4. anonymous

why don't you just use (4 +/- sqrt(20)i)/6

5. anonymous

huh?? O.o

6. zzr0ck3r

$$3x^2-4x+3=0\implies \dfrac{-(-4)\pm \sqrt{(-4)^2-4(3)(3)}}{2(3)}=\dfrac{4\pm\sqrt{16-36}}{6}\\=\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2\pm\sqrt{5}i}{3}$$

7. anonymous

Sorry.. Im assuming the 2 is from 4... But im a bit confused by the rest

8. anonymous

And shouldn't the i be before square root 5???

9. zzr0ck3r

the order of i does not matter, and I will show some more steps

10. zzr0ck3r

$$\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2(2+\sqrt{5}i)}{2*3}=(\dfrac{2}{2})\dfrac{2+\sqrt{5}i}{3}=(\dfrac{\cancel{2}}{\cancel{2}})\dfrac{2+\sqrt{5}i}{3}=\dfrac{2\pm\sqrt{5}i}{3}=\dfrac{2\pm i\sqrt{5}}{3}$$

11. anonymous

What is the last one??

12. zzr0ck3r

$$\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2(2+\sqrt{5}i)}{2*3}=(\dfrac{2}{2})\dfrac{2+\sqrt{5}i}{3}=(\dfrac{\cancel{2}}{\cancel{2}})\dfrac{2+\sqrt{5}i}{3}=\\\dfrac{2\pm\sqrt{5}i}{3}=\dfrac{2\pm i\sqrt{5}}{3}$$