k^2=41

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Solution for k^2=41 equation:



k^2=41
We move all terms to the left:
k^2-(41)=0
a = 1; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·1·(-41)
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{41}}{2*1}=\frac{0-2\sqrt{41}}{2} =-\frac{2\sqrt{41}}{2} =-\sqrt{41} $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{41}}{2*1}=\frac{0+2\sqrt{41}}{2} =\frac{2\sqrt{41}}{2} =\sqrt{41} $

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