(2x+4)(x)=16^2

Solution for (2x+4)(x)=16^2 equation:

(2x+4)(x)=16^2

We move all terms to the left:

(2x+4)(x)-(16^2)=0

We add all the numbers together, and all the variables

(2x+4)x-256=0

We multiply parentheses

2x^2+4x-256=0

a = 2; b = 4; c = -256;
Δ = b2-4ac
Δ = 42-4·2·(-256)
Δ = 2064
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:

$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{129}}{2*2}=\frac{-4-4\sqrt{129}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{129}}{2*2}=\frac{-4+4\sqrt{129}}{4}
$