p^2+2=21

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Solution for p^2+2=21 equation:



p^2+2=21
We move all terms to the left:
p^2+2-(21)=0
We add all the numbers together, and all the variables
p^2-19=0
a = 1; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·1·(-19)
Δ = 76
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:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{19}}{2*1}=\frac{0-2\sqrt{19}}{2} =-\frac{2\sqrt{19}}{2} =-\sqrt{19} $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{19}}{2*1}=\frac{0+2\sqrt{19}}{2} =\frac{2\sqrt{19}}{2} =\sqrt{19} $

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