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2y(3y+1)=3y+1
We move all terms to the left:
2y(3y+1)-(3y+1)=0
We multiply parentheses
6y^2+2y-(3y+1)=0
We get rid of parentheses
6y^2+2y-3y-1=0
We add all the numbers together, and all the variables
6y^2-1y-1=0
a = 6; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·6·(-1)
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*6}=\frac{-4}{12} =-1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*6}=\frac{6}{12} =1/2 $
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