Question

i need help finding the solutions

Answer 1

must use quadratic equation or complete the square

quadratic formula
for
ax^2+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

0=6x^2-2x+5
a=6
b=-2
c=5

x=$\frac{-(-2)+/- \sqrt{(-2)^2-4(6)(5)} }{2(6)}$
x=$\frac{2+/- \sqrt{4-120} }{12}$
x=$\frac{2+/- \sqrt{-116} }{12}$
x=$\frac{2+/- 2i\sqrt{29} }{12}$
x=$\frac{1+/- i\sqrt{29} }{6}$

x=$\frac{1+ i\sqrt{29} }{6}$ or x=$\frac{1- i\sqrt{29} }{6}$

Answer 2

6x² - 2x + 5 = 0

this isn't easy or pretty to factor, nor does it factor evenly, so you'll want to use the quadratic formula. the formatting of it will look a little strange on here, so i encourage you to check out your textbook or search "quadratic formula" if you aren't familiar. regardless:

(-b ± √b² - 4ac)/2a

in your equation, 6x² - 2x + 5 = 0, a is 6, -2 is b, 5 is c.

(-(-2) ± √(-2)² - 4(6)(5))/(2(6))

simplified: (2 ± √-116)/12

i started working this out further, but it gets really messy and it's super difficult to explain through words. if you solve it through with the quadratic formula, you'll get your answers, but it's a really rare and confusing case--the result isn't pretty. when i entered it into my calculator to get the zeroes i got (1/6) + 0.8975i and (1/6) - 0.8975i? so if you're getting something along those lines, you're good.