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-0.6x^2+3.6x=0
a = -0.6; b = 3.6; c = 0;
Δ = b2-4ac
Δ = 3.62-4·(-0.6)·0
Δ = 12.96
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{-(3.6)-\sqrt{12.96}}{2*-0.6}=\frac{-3.6-\sqrt{12.96}}{-1.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.6)+\sqrt{12.96}}{2*-0.6}=\frac{-3.6+\sqrt{12.96}}{-1.2} $
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