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0.05x^2+2.2x=0
a = 0.05; b = 2.2; c = 0;
Δ = b2-4ac
Δ = 2.22-4·0.05·0
Δ = 4.84
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{-(2.2)-\sqrt{4.84}}{2*0.05}=\frac{-2.2-\sqrt{4.84}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.2)+\sqrt{4.84}}{2*0.05}=\frac{-2.2+\sqrt{4.84}}{0.1} $
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