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5y^2+13y=3
We move all terms to the left:
5y^2+13y-(3)=0
a = 5; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·5·(-3)
Δ = 229
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{229}}{2*5}=\frac{-13-\sqrt{229}}{10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{229}}{2*5}=\frac{-13+\sqrt{229}}{10} $
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