Answer:
The zeros are [tex]\sqrt{3} + 2\ and \sqrt-{3} + 2[/tex]
Step-by-step explanation:
[tex]x = \frac{-b +- \sqrt{b^2 - 4ac} \\}{2a}[/tex]
Use the quadratic formula
Then plug in all the numbers
[tex]x = \frac{4 +- \sqrt{(-4)^2 - 4(1)(1)} \\}{2(1)}[/tex]
Then solve (-4)^2
[tex]x = \frac{4 +- \sqrt{16 - 4(1)(1)} \\}{2(1)}[/tex]
Then solve everything inside the radical the radical
[tex]\sqrt{12} = 2\sqrt{3}[/tex]
Then divide and simplify the radical. After that, you will get your final answer.
[tex]x = \frac{4 +- 2\sqrt{3)} \\}{2} = 2 + or - \sqrt{3}[/tex]
So your final answer is [tex]\sqrt{3} + 2\ and \sqrt-{3} + 2[/tex]
Hope it helped! My answer is expert verified.
