(2x+4)(2x+1)=6

Solution for (2x+4)(2x+1)=6 equation:

(2x+4)(2x+1)=6

We move all terms to the left:

(2x+4)(2x+1)-(6)=0

We multiply parentheses ..

(+4x^2+2x+8x+4)-6=0

We get rid of parentheses

4x^2+2x+8x+4-6=0

We add all the numbers together, and all the variables

4x^2+10x-2=0

a = 4; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·4·(-2)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*4}=\frac{-10-2\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*4}=\frac{-10+2\sqrt{33}}{8}
$