How do you evaluate #(8+ \sqrt { 3^ { 2} + 2\cdot 3+ 1})/-4#?

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Answer 1 · 2018-08-04T06:53:45

#-3#

#(8 + sqrt(3^2 + 2 * 3 + 1))/-4#

#=(8 + sqrt(9 + 6 + 1))/-4#

#=(8 + sqrt(16))/-4#

#=(8+4)/-4#

#=12/-4#

#=-3#

Hope this helps!