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

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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} $

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