3(x-2)+4(2-3x)=2X(3x+8)

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Solution for 3(x-2)+4(2-3x)=2X(3x+8) equation:



3(x-2)+4(2-3x)=2x(3x+8)
We move all terms to the left:
3(x-2)+4(2-3x)-(2x(3x+8))=0
We add all the numbers together, and all the variables
3(x-2)+4(-3x+2)-(2x(3x+8))=0
We multiply parentheses
3x-12x-(2x(3x+8))-6+8=0
We calculate terms in parentheses: -(2x(3x+8)), so:
2x(3x+8)
We multiply parentheses
6x^2+16x
Back to the equation:
-(6x^2+16x)
We add all the numbers together, and all the variables
-9x-(6x^2+16x)+2=0
We get rid of parentheses
-6x^2-9x-16x+2=0
We add all the numbers together, and all the variables
-6x^2-25x+2=0
a = -6; b = -25; c = +2;
Δ = b2-4ac
Δ = -252-4·(-6)·2
Δ = 673
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{673}}{2*-6}=\frac{25-\sqrt{673}}{-12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{673}}{2*-6}=\frac{25+\sqrt{673}}{-12} $

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