How do you solve #x^2+10x=13# using the quadratic equation?

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Answer 1 · 2017-10-25T10:49:27

Solution : #x= -5+-sqrt38 #

#x^2+10x=13 or x^2+10x-13=0#. Comparing wih

standard equation # ax^2+bx+c=0# we get here

#a=1 ,b=10 ,c=-13#. Discriminant D= b^2-4ac=100+52 =152#

Discriminant is positive , so it has two real roots .

Quadratic formula: #x= (-b+-sqrtD)/(2a) #

#:. x= (-10+-sqrt152)/(2*1) or x= (-10+-sqrt152)/2# or

#x= -5+-(cancel2sqrt38)/cancel2 or x= -5+-sqrt38 #

Solution : #x= -5+-sqrt38 # [Ans]