3y=2(2+2y)3y=4+4y-y=4y=

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Solution for 3y=2(2+2y)3y=4+4y-y=4y= equation:



3y=2(2+2y)3y=4+4y-y=4y=
We move all terms to the left:
3y-(2(2+2y)3y)=0
We add all the numbers together, and all the variables
3y-(2(2y+2)3y)=0
We calculate terms in parentheses: -(2(2y+2)3y), so:
2(2y+2)3y
We multiply parentheses
12y^2+12y
Back to the equation:
-(12y^2+12y)
We get rid of parentheses
-12y^2+3y-12y=0
We add all the numbers together, and all the variables
-12y^2-9y=0
a = -12; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-12)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-12}=\frac{0}{-24} =0 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-12}=\frac{18}{-24} =-3/4 $

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