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(p-3)(p+2)=14
We move all terms to the left:
(p-3)(p+2)-(14)=0
We multiply parentheses ..
(+p^2+2p-3p-6)-14=0
We get rid of parentheses
p^2+2p-3p-6-14=0
We add all the numbers together, and all the variables
p^2-1p-20=0
a = 1; b = -1; c = -20;
Δ = b2-4ac
Δ = -12-4·1·(-20)
Δ = 81
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{81}=9$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*1}=\frac{-8}{2} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*1}=\frac{10}{2} =5 $
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