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2p^2+6=62
We move all terms to the left:
2p^2+6-(62)=0
We add all the numbers together, and all the variables
2p^2-56=0
a = 2; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·2·(-56)
Δ = 448
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*2}=\frac{0-8\sqrt{7}}{4} =-\frac{8\sqrt{7}}{4} =-2\sqrt{7} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*2}=\frac{0+8\sqrt{7}}{4} =\frac{8\sqrt{7}}{4} =2\sqrt{7} $
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