3a^2+4+3a+2=2a^2+8a

Solution for 3a^2+4+3a+2=2a^2+8a equation:

3a^2+4+3a+2=2a^2+8a

We move all terms to the left:

3a^2+4+3a+2-(2a^2+8a)=0

We add all the numbers together, and all the variables

3a^2+3a-(2a^2+8a)+6=0

We get rid of parentheses

3a^2-2a^2+3a-8a+6=0

We add all the numbers together, and all the variables

a^2-5a+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$