Question

Discuss the nature of the roots x^2+6x=3

Answer 1

$x^2+6x=3\\ x^2+6x+9-9=3\\ (x+3)^2=12\\ x+3=\sqrt{12} \vee x+3=-\sqrt{12}\\ x=-3+\sqrt{12} \vee x=-3-\sqrt{12}\\ x=-3+2\sqrt{3} \vee x=-3-2\sqrt{3}$

Answer 2

$x^{2} +6x=3$
$x^{2} +6x-3=0$
using quadratic formula --> $x= \frac{-6± \sqrt{6^{2}-4 *1*-3} }{2*1}$ 
$x=-3+2 \sqrt{3}$ 
AND
$x=-3-2 \sqrt{3}$