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x^2+0.6x-0.9=0
a = 1; b = 0.6; c = -0.9;
Δ = b2-4ac
Δ = 0.62-4·1·(-0.9)
Δ = 3.96
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{-(0.6)-\sqrt{3.96}}{2*1}=\frac{-0.6-\sqrt{3.96}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{3.96}}{2*1}=\frac{-0.6+\sqrt{3.96}}{2} $
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