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y^+2y+1+3y(6y+1)=19y^
We move all terms to the left:
y^+2y+1+3y(6y+1)-(19y^)=0
We add all the numbers together, and all the variables
-16y+3y(6y+1)+1=0
We multiply parentheses
18y^2-16y+3y+1=0
We add all the numbers together, and all the variables
18y^2-13y+1=0
a = 18; b = -13; c = +1;
Δ = b2-4ac
Δ = -132-4·18·1
Δ = 97
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{97}}{2*18}=\frac{13-\sqrt{97}}{36} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{97}}{2*18}=\frac{13+\sqrt{97}}{36} $
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