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5x^2+0.6x=0
a = 5; b = 0.6; c = 0;
Δ = b2-4ac
Δ = 0.62-4·5·0
Δ = 0.36
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{-(0.6)-\sqrt{0.36}}{2*5}=\frac{-0.6-\sqrt{0.36}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{0.36}}{2*5}=\frac{-0.6+\sqrt{0.36}}{10} $
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