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x^2+2x-0.6=0
a = 1; b = 2; c = -0.6;
Δ = b2-4ac
Δ = 22-4·1·(-0.6)
Δ = 6.4
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{-(2)-\sqrt{6.4}}{2*1}=\frac{-2-\sqrt{6.4}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{6.4}}{2*1}=\frac{-2+\sqrt{6.4}}{2} $
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