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y^2-49y+180=0
a = 1; b = -49; c = +180;
Δ = b2-4ac
Δ = -492-4·1·180
Δ = 1681
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{1681}=41$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-41}{2*1}=\frac{8}{2} =4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+41}{2*1}=\frac{90}{2} =45 $
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