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5a(4a^-1)=0
We multiply parentheses
20a^2-5a=0
a = 20; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·20·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*20}=\frac{0}{40} =0 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*20}=\frac{10}{40} =1/4 $
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