4=3(i+1)4i+6

Solution for 4=3(i+1)4i+6 equation:

4=3(i+1)4i+6

We move all terms to the left:

4-(3(i+1)4i+6)=0


We calculate terms in parentheses: -(3(i+1)4i+6), so:

3(i+1)4i+6

We multiply parentheses

12i^2+12i+6

Back to the equation:

-(12i^2+12i+6)


We get rid of parentheses

-12i^2-12i-6+4=0

We add all the numbers together, and all the variables

-12i^2-12i-2=0

a = -12; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·(-12)·(-2)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*-12}=\frac{12-4\sqrt{3}}{-24}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*-12}=\frac{12+4\sqrt{3}}{-24}
$