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25y-5y^2=0
a = -5; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-5)·0
Δ = 625
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{625}=25$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-5}=\frac{-50}{-10} =+5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-5}=\frac{0}{-10} =0 $
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