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2(2y+2)=4y(3y+1)
We move all terms to the left:
2(2y+2)-(4y(3y+1))=0
We multiply parentheses
4y-(4y(3y+1))+4=0
We calculate terms in parentheses: -(4y(3y+1)), so:We get rid of parentheses
4y(3y+1)
We multiply parentheses
12y^2+4y
Back to the equation:
-(12y^2+4y)
-12y^2+4y-4y+4=0
We add all the numbers together, and all the variables
-12y^2+4=0
a = -12; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-12)·4
Δ = 192
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-12}=\frac{0-8\sqrt{3}}{-24} =-\frac{8\sqrt{3}}{-24} =-\frac{\sqrt{3}}{-3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-12}=\frac{0+8\sqrt{3}}{-24} =\frac{8\sqrt{3}}{-24} =\frac{\sqrt{3}}{-3} $
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