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p^2-15p+26=0
a = 1; b = -15; c = +26;
Δ = b2-4ac
Δ = -152-4·1·26
Δ = 121
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-11}{2*1}=\frac{4}{2} =2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+11}{2*1}=\frac{26}{2} =13 $
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