4a(3a+2)-4a=48

Solution for 4a(3a+2)-4a=48 equation:

4a(3a+2)-4a=48

We move all terms to the left:

4a(3a+2)-4a-(48)=0

We add all the numbers together, and all the variables

-4a+4a(3a+2)-48=0

We multiply parentheses

12a^2-4a+8a-48=0

We add all the numbers together, and all the variables

12a^2+4a-48=0

a = 12; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·12·(-48)
Δ = 2320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{145}}{2*12}=\frac{-4-4\sqrt{145}}{24}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{145}}{2*12}=\frac{-4+4\sqrt{145}}{24}
$