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5y-25y^2=0
a = -25; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-25)·0
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-25}=\frac{-10}{-50} =1/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-25}=\frac{0}{-50} =0 $
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