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3y^2+2y=184
We move all terms to the left:
3y^2+2y-(184)=0
a = 3; b = 2; c = -184;
Δ = b2-4ac
Δ = 22-4·3·(-184)
Δ = 2212
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2212}=\sqrt{4*553}=\sqrt{4}*\sqrt{553}=2\sqrt{553}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{553}}{2*3}=\frac{-2-2\sqrt{553}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{553}}{2*3}=\frac{-2+2\sqrt{553}}{6} $
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