3x(2+4x)=20

Solution for 3x(2+4x)=20 equation:

3x(2+4x)=20

We move all terms to the left:

3x(2+4x)-(20)=0

We add all the numbers together, and all the variables

3x(4x+2)-20=0

We multiply parentheses

12x^2+6x-20=0

a = 12; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·12·(-20)
Δ = 996
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:

$\sqrt{\Delta}=\sqrt{996}=\sqrt{4*249}=\sqrt{4}*\sqrt{249}=2\sqrt{249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{249}}{2*12}=\frac{-6-2\sqrt{249}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{249}}{2*12}=\frac{-6+2\sqrt{249}}{24}
$