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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+16a^2+8a+8a+4)-6=0

We get rid of parentheses

16a^2+8a+8a+4-6=0

We add all the numbers together, and all the variables

16a^2+16a-2=0

a = 16; b = 16; c = -2;
Δ = b2-4ac
Δ = 162-4·16·(-2)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{6}}{2*16}=\frac{-16-8\sqrt{6}}{32}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{6}}{2*16}=\frac{-16+8\sqrt{6}}{32}
$