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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-6=0

a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2}
=-\frac{2\sqrt{6}}{2}
=-\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2}
=\frac{2\sqrt{6}}{2}
=\sqrt{6}
$