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y^2-25y+100=0
a = 1; b = -25; c = +100;
Δ = b2-4ac
Δ = -252-4·1·100
Δ = 225
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{225}=15$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15}{2*1}=\frac{10}{2} =5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15}{2*1}=\frac{40}{2} =20 $
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