Respuesta :

fieryanswererft

answer:

B and C are correct.

explanation:

[tex]\hookrightarrow \sf x^2-5x+5[/tex]  where ax² + bx + c

\\ Thus we can determine a = 1, b = -5, c = 5 \\

using quadratic equation:

[tex]\rightarrow \sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ when \ \ ax^2 + bx + c = 0[/tex]

[tex]\hookrightarrow \sf x_{1,\:2}=\dfrac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}[/tex]

[tex]\hookrightarrow \sf x=\dfrac{5+\sqrt{5}}{2},\:x=\dfrac{5-\sqrt{5}}{2}[/tex]

semsee45

Answer:

B and C

Step-by-step explanation:

[tex]x^2-5x+5=0[/tex]

⇒ a = 1, b = -5, c = 5

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{5\pm\sqrt{(-5)^2-4\cdot 1 \cdot 5} }{2\cdot 1}[/tex]

[tex]\implies x=\dfrac{5\pm\sqrt{5}}{2}[/tex]

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