Answer :
Step-by-step explanation:
Hey there!!
Given,
[tex] {x}^{2} + 8x - 5 = 0[/tex]
Comparing it with ax^2+bx+c =0 we get,
a= 1, b= 8, c= -5.
Using formula,
[tex]x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]
Keep all values,
[tex]x = \frac{ - 8 + - \sqrt{ {8}^{2} - 4 \times 1 \times ( - 5)} }{2 \times 1} [/tex]
[tex]x = \frac{ - 8 + - \sqrt{64 + 20} }{2} [/tex]
[tex]x = \frac{ - 8 + - \sqrt{84} }{2} [/tex]
[tex]x = \frac{ - 8 + - 2 \sqrt{21} }{2} [/tex]
Taking negative,
[tex]x = \frac{ - 8 - 2 \sqrt{21} }{2} [/tex]
[tex]x = \frac{ - 2(4 + \sqrt{21}) }{2} [/tex]
[tex]x = 4 + \sqrt{21} [/tex]
Similarly, taking positive,
[tex]x = \frac{ - 8 + 2 \sqrt{21} }{2} [/tex]
[tex]x = \frac{ - 2(4 - \sqrt{21} ) }{2} [/tex]
[tex]x = 4 - \sqrt{21} [/tex]
Therefore, x= (4 + root 21, 4 - root 21).
Hope it helps.....