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2p^2=100
We move all terms to the left:
2p^2-(100)=0
a = 2; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·2·(-100)
Δ = 800
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*2}=\frac{0-20\sqrt{2}}{4} =-\frac{20\sqrt{2}}{4} =-5\sqrt{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*2}=\frac{0+20\sqrt{2}}{4} =\frac{20\sqrt{2}}{4} =5\sqrt{2} $
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