j2+9=11

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Solution for j2+9=11 equation:



j2+9=11
We move all terms to the left:
j2+9-(11)=0
We add all the numbers together, and all the variables
j^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:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$j_{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} $
$j_{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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