Respuesta :
To solve a quadratic equation like [tex] ax^2+bx+c=0 [/tex], you can use the quadratic formula
[tex] x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
In your case, [tex] a = 1, b = c = 3 [/tex], so the formula becomes
[tex] x_{1,2} = \cfrac{-3\pm\sqrt{(-3)^2-4\cdot 1\cdot 3}}{2\cdot 1} [/tex]
We can simplify the expression:
[tex] x_{1,2} = \cfrac{-3\pm\sqrt{9-12}}{2} = \cfrac{-3\pm\sqrt{-3}}{2}[/tex]
Since -3 is negative, its square root is computed as
[tex] \sqrt{-3} = \sqrt{-1\cdot 3} = \sqrt{-1}\sqrt{3} = \sqrt{3}i [/tex]
So, the solutions are
[tex] x = \cfrac{-3+i\sqrt{3}}{2} \text{ or } x = \cfrac{-3-i\sqrt{3}}{2} [/tex]
Answer:
Step-by-step explanation:
The answer is C. `(-3 stackrel(+)(-) isqrt(3))/(2)`
