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4x^2-1.6x=0
a = 4; b = -1.6; c = 0;
Δ = b2-4ac
Δ = -1.62-4·4·0
Δ = 2.56
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{-(-1.6)-\sqrt{2.56}}{2*4}=\frac{1.6-\sqrt{2.56}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.6)+\sqrt{2.56}}{2*4}=\frac{1.6+\sqrt{2.56}}{8} $
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