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y^2+2y-440=0
a = 1; b = 2; c = -440;
Δ = b2-4ac
Δ = 22-4·1·(-440)
Δ = 1764
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-42}{2*1}=\frac{-44}{2} =-22 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+42}{2*1}=\frac{40}{2} =20 $
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