Respuesta :

Answer:

[tex]x=\frac{1+\sqrt{41}}{5},\\x=\frac{1-\sqrt{41}}{5}[/tex]

Step-by-step explanation:

The quadratic formula states that the solutions for a quadratic is standard form [tex]ax^2+bx+c[/tex] are equal to [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].

In [tex]5x^2-2x-8=0[/tex], we can assign the values:

  • [tex]a[/tex] of 5
  • [tex]b[/tex] of -2
  • [tex]c[/tex] of -8

Thus, we have:

[tex]x=\frac{-(-2)\pm \sqrt{(-2)^2-4(5)(-8)}}{2(5)},\\x=\frac{2\pm \sqrt{164}}{10},\\\begin{cases}x=\frac{2+ \sqrt{164}}{10}, x=\frac{1}{5}+\frac{\sqrt{41}}{5}=\frac{1+\sqrt{41}}{5}\\x=\frac{2- \sqrt{164}}{10}, x=\frac{1}{5}-\frac{\sqrt{41}}{5}=\frac{1-\sqrt{41}}{5}\end{cases}[/tex]

Answer:

(1+√41)/5, (1-√41)/5

Step-by-step explanation:

quadratic formula is (-b±√(b^2-4ac))/2a

in this equation,

a = 5

b = -2

c = -8

plug in the values

(2±√(4 - 4(5)(-8))/10

(2±√(4 + 160)/10

(2±√(164)/10

(2±2√(41))/10

1. (2+2√41)/10

(1+√41)/5

2. (2-2√41)/10

(1-√41)/5