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y^2+13y-28=0
a = 1; b = 13; c = -28;
Δ = b2-4ac
Δ = 132-4·1·(-28)
Δ = 281
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{-(13)-\sqrt{281}}{2*1}=\frac{-13-\sqrt{281}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{281}}{2*1}=\frac{-13+\sqrt{281}}{2} $
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