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4y(2y+1)=3y
We move all terms to the left:
4y(2y+1)-(3y)=0
We add all the numbers together, and all the variables
-3y+4y(2y+1)=0
We multiply parentheses
8y^2-3y+4y=0
We add all the numbers together, and all the variables
8y^2+y=0
a = 8; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·8·0
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*8}=\frac{-2}{16} =-1/8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*8}=\frac{0}{16} =0 $
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