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20x^2+3x+0.01=0
a = 20; b = 3; c = +0.01;
Δ = b2-4ac
Δ = 32-4·20·0.01
Δ = 8.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{8.2}}{2*20}=\frac{-3-\sqrt{8.2}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{8.2}}{2*20}=\frac{-3+\sqrt{8.2}}{40} $
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