Answer:
[tex]x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}[/tex]
Step-by-step explanation:
[tex]x^{2}[/tex]+6x-5=0
we divide the coefficient of the X by half :
in this case: 6/2 = 3 , then we do the following
The result obtained is raised to square power: 3^2=9
we sum and subtract by 9 to maintain the balance of the equation:
[tex]x^{2}[/tex]+6x+9-9-5=0
we have:
[tex](x+3)^{2}[/tex]-9-5=0
[tex](x+3)^{2}[/tex] = 14
lets apply square root on both sides of the equation:
[tex]\sqrt{(x+3)^{2}}=\sqrt{14}[/tex]
we know:
[tex]\sqrt{a^{2}} = abs(a)[/tex]
so we have:
abs(x+3)=[tex]\sqrt{14}[/tex]
from where two solutions are obtained
[tex]x_{1} + 3 =\sqrt{14} \\\\x_{2} + 3 =-\sqrt{14}[/tex]
finally we have:
[tex]x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}[/tex]
