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y^2+y-0.09=0
a = 1; b = 1; c = -0.09;
Δ = b2-4ac
Δ = 12-4·1·(-0.09)
Δ = 1.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1.36}}{2*1}=\frac{-1-\sqrt{1.36}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.36}}{2*1}=\frac{-1+\sqrt{1.36}}{2} $
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