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p^2-13p-90=0
a = 1; b = -13; c = -90;
Δ = b2-4ac
Δ = -132-4·1·(-90)
Δ = 529
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{529}=23$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*1}=\frac{-10}{2} =-5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*1}=\frac{36}{2} =18 $
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