How do you solve #x^2 + 5 = 41#?

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Answer 1 · 2018-05-13T05:45:00

#x=6#

#x^2+5=41#

#x^2=36#

#pm6^2=36#

#x=pm6#

hope this helps

Answer 2 · 2018-05-13T06:50:36

#x=pm6#

#x^2+5=41#

#x^2+5color(red)(-5)=41color(red)(-5)#

#x^2=36#

#sqrt(x^2)=sqrt36#

#x=pm6#

We can graph both the right and left hand sides:

graph{(y-x^2-5)(y-0x-41)=0 [-11.31, 12.99, 32.25, 44.41]}