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p^2=31-10p
We move all terms to the left:
p^2-(31-10p)=0
We add all the numbers together, and all the variables
p^2-(-10p+31)=0
We get rid of parentheses
p^2+10p-31=0
a = 1; b = 10; c = -31;
Δ = b2-4ac
Δ = 102-4·1·(-31)
Δ = 224
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{14}}{2*1}=\frac{-10-4\sqrt{14}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{14}}{2*1}=\frac{-10+4\sqrt{14}}{2} $
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