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x^2+2x-0.28=0
a = 1; b = 2; c = -0.28;
Δ = b2-4ac
Δ = 22-4·1·(-0.28)
Δ = 5.12
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{5.12}}{2*1}=\frac{-2-\sqrt{5.12}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{5.12}}{2*1}=\frac{-2+\sqrt{5.12}}{2} $
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