(2x)(2x+2)=168

Solution for (2x)(2x+2)=168 equation:

(2x)(2x+2)=168

We move all terms to the left:

(2x)(2x+2)-(168)=0

We multiply parentheses

4x^2+4x-168=0

a = 4; b = 4; c = -168;
Δ = b2-4ac
Δ = 42-4·4·(-168)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-52}{2*4}=\frac{-56}{8}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+52}{2*4}=\frac{48}{8}
=6
$