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1-0.5x+0.06x^2=0
a = 0.06; b = -0.5; c = +1;
Δ = b2-4ac
Δ = -0.52-4·0.06·1
Δ = 0.01
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{0.01}}{2*0.06}=\frac{0.5-\sqrt{0.01}}{0.12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{0.01}}{2*0.06}=\frac{0.5+\sqrt{0.01}}{0.12} $
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