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