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11y^2+7+18y=0
a = 11; b = 18; c = +7;
Δ = b2-4ac
Δ = 182-4·11·7
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4}{2*11}=\frac{-22}{22} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4}{2*11}=\frac{-14}{22} =-7/11 $
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