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y^2+28y+96=0
a = 1; b = 28; c = +96;
Δ = b2-4ac
Δ = 282-4·1·96
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*1}=\frac{-48}{2} =-24 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*1}=\frac{-8}{2} =-4 $
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