How do you solve #x^2 - 5x = 1# using the quadratic formula?

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Answer 1 · 2015-12-21T06:01:20

#x=(5+sqrt 29)/2#

#x=(5-sqrt29)/2#

#x^2-5x=1#

Subtract #1# from both sides.

#x^2-5x-1=0# is a quadratic equation in standard form, #ax^2+bx+c#, where #a=1, b=-5, and c=-1#.

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

Substitute the values for #a, b, and c# into the formula.

#x=(-(-5)+-sqrt(-5^2-4*1*-1))/(2*1)#

Simplify.

#x=(5+-sqrt(25-(-4)))/2#

#x=(5+-sqrt(25+4))/2#

#x=(5+-sqrt29)/2#

Solve for #x#

#x=(5+sqrt 29)/2#

#x=(5-sqrt29)/2#