4a-3(a-2)=2(3a-2)4a-3a+6=6a-4

Solution for 4a-3(a-2)=2(3a-2)4a-3a+6=6a-4 equation:

4a-3(a-2)=2(3a-2)4a-3a+6=6a-4

We move all terms to the left:

4a-3(a-2)-(2(3a-2)4a-3a+6)=0

We multiply parentheses

4a-3a-(2(3a-2)4a-3a+6)+6=0


We calculate terms in parentheses: -(2(3a-2)4a-3a+6), so:

2(3a-2)4a-3a+6

We add all the numbers together, and all the variables

-3a+2(3a-2)4a+6

We multiply parentheses

24a^2-3a-16a+6

We add all the numbers together, and all the variables

24a^2-19a+6

Back to the equation:

-(24a^2-19a+6)


We add all the numbers together, and all the variables

a-(24a^2-19a+6)+6=0

We get rid of parentheses

-24a^2+a+19a-6+6=0

We add all the numbers together, and all the variables

-24a^2+20a=0

a = -24; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-24)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-24}=\frac{-40}{-48}
=5/6
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-24}=\frac{0}{-48}
=0
$