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y^2-41y+348=0
a = 1; b = -41; c = +348;
Δ = b2-4ac
Δ = -412-4·1·348
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-17}{2*1}=\frac{24}{2} =12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+17}{2*1}=\frac{58}{2} =29 $
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