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1+0.5x-0.4x^2=0
a = -0.4; b = 0.5; c = +1;
Δ = b2-4ac
Δ = 0.52-4·(-0.4)·1
Δ = 1.85
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{-(0.5)-\sqrt{1.85}}{2*-0.4}=\frac{-0.5-\sqrt{1.85}}{-0.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{1.85}}{2*-0.4}=\frac{-0.5+\sqrt{1.85}}{-0.8} $
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