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x^2+1.6x+0.63=0
a = 1; b = 1.6; c = +0.63;
Δ = b2-4ac
Δ = 1.62-4·1·0.63
Δ = 0.04
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{0.04}}{2*1}=\frac{-1.6-\sqrt{0.04}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.6)+\sqrt{0.04}}{2*1}=\frac{-1.6+\sqrt{0.04}}{2} $
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