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-0.2x^2+x+5=0
a = -0.2; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-0.2)·5
Δ = 5
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)-\sqrt{5}}{2*-0.2}=\frac{-1-\sqrt{5}}{-0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5}}{2*-0.2}=\frac{-1+\sqrt{5}}{-0.4} $
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