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2p^2+11p-50=0
a = 2; b = 11; c = -50;
Δ = b2-4ac
Δ = 112-4·2·(-50)
Δ = 521
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{-(11)-\sqrt{521}}{2*2}=\frac{-11-\sqrt{521}}{4} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{521}}{2*2}=\frac{-11+\sqrt{521}}{4} $
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