Please show me how to find the acute angle between y=2x+4 and y=-3x+6 the answer is 45 degrees

One has a slope of m1 = 2 and the other has a slope of m2 = -3.

The formula is (m1-m2) / (1 + m1*m2) = 5/5 = 1

acrtan(1) = 45 degrees

If I understand the question correctly, you are looking for the angle between the two given lines.

y= 2x+4 is equivalent to the vector A = <1, 2>
y = -3x+6 is equivalent to the vector B = <1,-3>

The angle between the two vectors can be determined from the defintion of the scalar product of two vectors:
AdotB = |A||B|cosΘ
Θ = arccos[AdotB/(|A||B|)]
|A| = √(1²+2²) = √5
|B| = √(1²+3²) = √10
|A||B| = √50 = 5√2
AdotB = 1*1 + 2*(-3) = -5
Θ = arccos[-5/(5√2)] = -√2/2 = 135°
Since this is an obtuse angle, the acute angle desired is the supplement of the obtuse angle.

Θ = 180 - 135 = 45°

{2,-1} is a vector normal to y = 2x + 4
{3,1} is a vector normal to y = 6 - 3x
cos(t) = {2,-1}.{3,1} / (||{2,-1}|| ||{3,1}||)
= 5 / (√5 √10)
= 1/√2
t = π/4 or 45°

Answer: see above