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2y^2-23y+11=0
a = 2; b = -23; c = +11;
Δ = b2-4ac
Δ = -232-4·2·11
Δ = 441
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{441}=21$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-21}{2*2}=\frac{2}{4} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+21}{2*2}=\frac{44}{4} =11 $
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