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k^2=69
We move all terms to the left:
k^2-(69)=0
a = 1; b = 0; c = -69;
Δ = b2-4ac
Δ = 02-4·1·(-69)
Δ = 276
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{69}}{2*1}=\frac{0-2\sqrt{69}}{2} =-\frac{2\sqrt{69}}{2} =-\sqrt{69} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{69}}{2*1}=\frac{0+2\sqrt{69}}{2} =\frac{2\sqrt{69}}{2} =\sqrt{69} $
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