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k2=103
We move all terms to the left:
k2-(103)=0
We add all the numbers together, and all the variables
k^2-103=0
a = 1; b = 0; c = -103;
Δ = b2-4ac
Δ = 02-4·1·(-103)
Δ = 412
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{103}}{2*1}=\frac{0-2\sqrt{103}}{2} =-\frac{2\sqrt{103}}{2} =-\sqrt{103} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{103}}{2*1}=\frac{0+2\sqrt{103}}{2} =\frac{2\sqrt{103}}{2} =\sqrt{103} $
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