\Ans{$f(x)=3x^2-12x+2$
	
	At $x=3$
	$$f(3)=-7$$
	At $x=4$
	$$f(4)=2$$
	So, Roots between $3$ and $4$
	$$f'(x)=6x-12$$
	Now, by using newton-raphson method
	$$x_{n+1}=x_n-\frac{f(x)}{f'(x)}$$
	take $x_0=3.5$
	$$x_1=x_0-\frac{3{x_0}^2-12x_0+2}{6x_0-12}$$
	$1^{st}$ iteration:
	$$x_1=0.5-\frac{3(3.5)^2-12(3.5)+2}{6(3.5)-12}$$
	$$x_1=3.8611$$
	$2^{nd}$ iteration:
    $$x_2=3.8611-\frac{3(3.8611)^2-12(3.8611)+2}{6(3.8611)-12}$$
    $$x_2=3.8260$$
    $3^{rd}$ iteration:
    $$x_3=3.8260-\frac{3(3.8260)^2-12(3.8260)+2}{6(3.8260)-12}$$
    $$x_3=3.8257$$
    $4^{th}$ iteration:
    $$x_4=3.8257-\frac{3(3.8257)^2-12(3.8257)+2}{6(3.8257)-12}$$
    $$x_4=3.8257$$
    So, real root of equation is $3.8257$}