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p^2-24=2p
We move all terms to the left:
p^2-24-(2p)=0
a = 1; b = -2; c = -24;
Δ = b2-4ac
Δ = -22-4·1·(-24)
Δ = 100
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{100}=10$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*1}=\frac{-8}{2} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*1}=\frac{12}{2} =6 $
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