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0.8x^2+3.2x=0
a = 0.8; b = 3.2; c = 0;
Δ = b2-4ac
Δ = 3.22-4·0.8·0
Δ = 10.24
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{-(3.2)-\sqrt{10.24}}{2*0.8}=\frac{-3.2-\sqrt{10.24}}{1.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{10.24}}{2*0.8}=\frac{-3.2+\sqrt{10.24}}{1.6} $
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