Given,
x2−4x+3+x2−9=4x2−14x+6
⇒(x−3)(x−1)+(x−3)(x+3)=(x−3)(4x−2)
⇒(x−3)(x−1)+(x−3)(x+3)−(x−3)(4x−2)=0
⇒(x−3)=0or [(x−1)+(x+3)−(4x−2)]=0
So, x=3 is one solution and also x≥3
Now solving [(x−1)+(x+3)−(4x−2)]=0
⇒(x−1)+(x+3)=(4x−2)
Now squaring both side we get,
⇒(x−1)+(x+3)+2(x−1)(x+3)=(4x−2)
⇒2(x−1)(x+3)=(2x−4)
⇒(x−1)(x+3)=(x−2)2
⇒x2+2x−3=x2−4x+4
⇒6x=7⇒x=67 which does not satisfy the given expression and x≥3
So, there is only one solution which is x=3