Answer:
Two values of x that are :
[tex]x=-\sqrt{7}-1[/tex]
[tex]x=\sqrt{7}-1[/tex]
Step-by-step explanation:
Given= [tex]x^2+2x-6=0[/tex]
To find = Two values of x
Solution :
[tex]x^2+2x-6=0[/tex]
Adding and subtracting 1 in the equation :
[tex]x^2+2x+1-1-6=0[/tex]
[tex](x^2+2\times x\times 1+1^2)-1-6=0[/tex]
[tex](x+1)^2-7=0[/tex]
[tex](x+1)^2=7[/tex]
[tex](x+1)=\pm \sqrt{7}[/tex]
Two values of x that are :
[tex]x=-\sqrt{7}-1[/tex]
[tex]x=\sqrt{7}-1[/tex]
