How do you solve #x^2-5x= -9# using the quadratic formula?

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Answer 1 · 2016-10-07T20:09:02

2 complex roots, #x=(5+-3i)/(2)#

Original Equation

#x^2-5x=-9 => x^2-5x+9=0#

Standard Form of a quadratic equation

#ax^2+bx+c=0#

#a=1#
#b=-5#
#c=9#

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

#x=(-(-5)+-sqrt((-5)^2-4(1)(9)))/(2(1))#

#x=(5+-sqrt(25-36))/(2)#

#x=(5+-sqrt(-9))/(2)#

#x=(5+-3i)/(2)#

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