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2p^2=154
We move all terms to the left:
2p^2-(154)=0
a = 2; b = 0; c = -154;
Δ = b2-4ac
Δ = 02-4·2·(-154)
Δ = 1232
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1232}=\sqrt{16*77}=\sqrt{16}*\sqrt{77}=4\sqrt{77}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{77}}{2*2}=\frac{0-4\sqrt{77}}{4} =-\frac{4\sqrt{77}}{4} =-\sqrt{77} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{77}}{2*2}=\frac{0+4\sqrt{77}}{4} =\frac{4\sqrt{77}}{4} =\sqrt{77} $
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