(4x+6x)(2x+4x)=126

Solution for (4x+6x)(2x+4x)=126 equation:

(4x+6x)(2x+4x)=126

We move all terms to the left:

(4x+6x)(2x+4x)-(126)=0

We add all the numbers together, and all the variables

(+10x)(+6x)-126=0

We multiply parentheses ..

(+60x^2)-126=0

We get rid of parentheses

60x^2-126=0

a = 60; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·60·(-126)
Δ = 30240
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:

$\sqrt{\Delta}=\sqrt{30240}=\sqrt{144*210}=\sqrt{144}*\sqrt{210}=12\sqrt{210}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{210}}{2*60}=\frac{0-12\sqrt{210}}{120}
=-\frac{12\sqrt{210}}{120}
=-\frac{\sqrt{210}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{210}}{2*60}=\frac{0+12\sqrt{210}}{120}
=\frac{12\sqrt{210}}{120}
=\frac{\sqrt{210}}{10}
$