Answer:
[tex]\frac{1}{2}[/tex]
Step-by-step explanation:
Quadratic formula goes as follows [tex]\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] and [tex]\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex], there is no plus or minus (+/- thingy) sign on here but use both plus and minus on the numerator where I switched the signs.
The equation you give is [tex]3x^2-3x+\frac{3}{4}[/tex], to make it nice, multiply everything by 4 to cancel out the fraction and end up with [tex]12x^2-12x+3[/tex]. So a=12, b=-12, and c=3. Use these values and plug them into quadratic formula.
For the first formula I wrote, [tex]\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex], you get [tex]\frac{-(-12)+\sqrt{(-12)^2-4(12)(3)} }{2(12)}[/tex]= [tex]\frac{1}{2}[/tex].
For the second formula, [tex]\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex], you get [tex]\frac{-(-12)-\sqrt{(-12)^2-4(12)(3)} }{2(12)}[/tex]= [tex]\frac{1}{2}[/tex].
Always do quadratic formula for both plus and minus to get both zeros. However, in this case, the answer still remained positive. So the zero of this would be just [tex]\frac{1}{2}[/tex].
