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0.01a^2+0.06a+0.01=0
a = 0.01; b = 0.06; c = +0.01;
Δ = b2-4ac
Δ = 0.062-4·0.01·0.01
Δ = 0.0032
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.06)-\sqrt{0.0032}}{2*0.01}=\frac{-0.06-\sqrt{0.0032}}{0.02} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.06)+\sqrt{0.0032}}{2*0.01}=\frac{-0.06+\sqrt{0.0032}}{0.02} $
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