Question 315287
Well that's certainly not the easy way.
You have two equations.
{{{y=3x^2+4x-1}}}
{{{y=-2x+b}}}
Look for the intersection point.
Set them equal to each other. 
{{{3x^2+4x-1=-2x+b}}}
{{{3x^2+6x-(1+b)=0}}}
Since the line is a tangent point, the curves only intersect at one point. 
Then x must be a double root and the equation must be a perfect square.
Complete the square to solve for both x and b.
{{{3(x^2+2x)-(1+b)=0}}}
{{{3(x^2+2x+1)-(1+b)=3}}}
{{{3(x+1)^2=1+b+3}}}
{{{3(x+1)^2=b+4=0}}}
The x coordinate for intersection is {{{x=-1}}}
and also {{{b=-4}}}
{{{highlight(y=-2x-4)}}}