How do you solve #(r-2)(r+6)=4# using the quadratic formula?

1 Answer
Jan 23, 2017

Answer:

Solution: #r=2.47(2dp) or r= -6.47(2dp)#

Explanation:

#(r-2)(r+6)=4 or r^2+4r-12=4 or r^2+4r-16=0 or (r+2)^2-4-16=0 or (r+2)^2=20 or r+2 = +-sqrt(20) or r+2=+-2sqrt5 :. r= -2+2sqrt5=2(-1+sqrt5)=2.47(2dp) =or r= -2-2sqrt5= 2(-1-sqrt5)=-6.47(2dp)#
Solution: #r=2.47(2dp) or r= -6.47(2dp)#

Using Quadratic formula:#(r=-b/(2a)+-sqrt(b^2-4ac)/(2a))#

#r^2+4r-16=0 ; a=1 ;b= 4 ;c=-16 ; b^2-4ac=80 ; r= -4/2+- sqrt80/2= -2+-2sqrt5:. r_1=2.47(2dp) or r_2=-6.47(2dp)#
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