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10y^2+13y+3=0
a = 10; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·10·3
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*10}=\frac{-20}{20} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*10}=\frac{-6}{20} =-3/10 $
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