How do you solve #x^2 - 4x -3 = 0#?

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Answer 1 · 2015-08-09T20:55:28

Use the quadratic formula to find #x = 2+-sqrt(7)#

#x^2-4x-3# is of the form #ax^2+bx+c#, with #a=1#, #b=-4# and #c=-3#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-4)^2-(4xx1xx-3) = 16+12 = 28#

#= 2^2*7#

This is positive but not a perfect square, so #x^2-4x-3 = 0# has two distinct irrational roots.

Use the quadratic formula to give the roots:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)#

#= (4+-2sqrt(7))/2 = 2+-sqrt(7)#