Answer:
we have added 9/4 to both sides of the equation.
Step-by-step explanation:
The given equation is :
[tex]x^2+3x=6[/tex]
Here, a = 1, b = 3 and c = 6
We should add (b/2)^2 to the both sides of the equation.
(b/2)^2 = (3/2)^2 = (9/4)
[tex]x^2+3x+\dfrac{9}{4}=6+\dfrac{9}{4}\\\\(x+\dfrac{3}{2})^2=\dfrac{33}{4}\\\\x+\dfrac{3}{2}=\pm \dfrac{\sqrt{33}}{2}\\\\x= +\dfrac{\sqrt{33}}{2}-\dfrac{3}{2}, -\dfrac{\sqrt{33}}{2}-\dfrac{3}{2}\\\\=1.37,-4.37[/tex]
Hence, we have added 9/4 to both sides of the equation.
