Answer:
x = 3 + √6 ; x = 3 - √6 ;
; 
Step-by-step explanation:
Relation given in the question:
(x² − 6x +3)(2x² − 4x − 7) = 0
Now,
for the above relation to be true the following condition must be followed:
Either (x² − 6x +3) = 0 ............(1)
or
(2x² − 4x − 7) = 0 ..........(2)
now considering the equation (1)
(x² − 6x +3) = 0
the roots can be found out as:

for the equation ax² + bx + c = 0
thus,
the roots are

or

or
and, x = 
or
and, x = 
or
x = 3 + √6 and x = 3 - √6
similarly for (2x² − 4x − 7) = 0.
we have
the roots are

or

or
and, x = 
or
and, x = 
or
and, x = 
or
and, 
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ;
; 
Answer:
Unknown answer so ye that mean I'm not sure about it
Step-by-step explanation:
Unknown answer so ye that mean I'm not sure about it
Eighteen tens < 21 × (5+5)
Hope this helps!
Using BEDMAS
a) 6+4(2+3)^2
6+4(5)^2 - Eliminate addition in bracket
6+4(25) - Execute the exponent
6+100 - Mulitplying
106 - Addition
b)(6+4)(2+3)
(10)(5) -Addition in BRACKETS done
50 - Mulipication
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Answer:
I think it would be 2 infinity