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13+5y-2y^2=0
a = -2; b = 5; c = +13;
Δ = b2-4ac
Δ = 52-4·(-2)·13
Δ = 129
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{-(5)-\sqrt{129}}{2*-2}=\frac{-5-\sqrt{129}}{-4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{129}}{2*-2}=\frac{-5+\sqrt{129}}{-4} $
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