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