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6y^2-19y+13=0
a = 6; b = -19; c = +13;
Δ = b2-4ac
Δ = -192-4·6·13
Δ = 49
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{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-7}{2*6}=\frac{12}{12} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+7}{2*6}=\frac{26}{12} =2+1/6 $
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