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0.6x^2+0.05x-0.05=0
a = 0.6; b = 0.05; c = -0.05;
Δ = b2-4ac
Δ = 0.052-4·0.6·(-0.05)
Δ = 0.1225
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{-(0.05)-\sqrt{0.1225}}{2*0.6}=\frac{-0.05-\sqrt{0.1225}}{1.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.05)+\sqrt{0.1225}}{2*0.6}=\frac{-0.05+\sqrt{0.1225}}{1.2} $
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