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-0.1x^2+0.5x+0.6=0
a = -0.1; b = 0.5; c = +0.6;
Δ = b2-4ac
Δ = 0.52-4·(-0.1)·0.6
Δ = 0.49
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{0.49}}{2*-0.1}=\frac{-0.5-\sqrt{0.49}}{-0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.49}}{2*-0.1}=\frac{-0.5+\sqrt{0.49}}{-0.2} $
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