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p2+1=3
We move all terms to the left:
p2+1-(3)=0
We add all the numbers together, and all the variables
p^2-2=0
a = 1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·1·(-2)
Δ = 8
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*1}=\frac{0-2\sqrt{2}}{2} =-\frac{2\sqrt{2}}{2} =-\sqrt{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*1}=\frac{0+2\sqrt{2}}{2} =\frac{2\sqrt{2}}{2} =\sqrt{2} $
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