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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(3(2x+4)x), so:

3(2x+4)x

We multiply parentheses

6x^2+12x

Back to the equation:

-(6x^2+12x)


We get rid of parentheses

-6x^2+2x-12x+20=0

We add all the numbers together, and all the variables

-6x^2-10x+20=0

a = -6; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·(-6)·20
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{145}}{2*-6}=\frac{10-2\sqrt{145}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{145}}{2*-6}=\frac{10+2\sqrt{145}}{-12}
$