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3p+63=p2
We move all terms to the left:
3p+63-(p2)=0
We add all the numbers together, and all the variables
-1p^2+3p+63=0
a = -1; b = 3; c = +63;
Δ = b2-4ac
Δ = 32-4·(-1)·63
Δ = 261
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{29}}{2*-1}=\frac{-3-3\sqrt{29}}{-2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{29}}{2*-1}=\frac{-3+3\sqrt{29}}{-2} $
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