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p^2+15p+11=0
a = 1; b = 15; c = +11;
Δ = b2-4ac
Δ = 152-4·1·11
Δ = 181
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{181}}{2*1}=\frac{-15-\sqrt{181}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{181}}{2*1}=\frac{-15+\sqrt{181}}{2} $
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