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p^2-18p+32=0
a = 1; b = -18; c = +32;
Δ = b2-4ac
Δ = -182-4·1·32
Δ = 196
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{196}=14$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-14}{2*1}=\frac{4}{2} =2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14}{2*1}=\frac{32}{2} =16 $
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