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11p^2+7p-16=0
a = 11; b = 7; c = -16;
Δ = b2-4ac
Δ = 72-4·11·(-16)
Δ = 753
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{-(7)-\sqrt{753}}{2*11}=\frac{-7-\sqrt{753}}{22} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{753}}{2*11}=\frac{-7+\sqrt{753}}{22} $
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