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k2=32
We move all terms to the left:
k2-(32)=0
We add all the numbers together, and all the variables
k^2-32=0
a = 1; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·1·(-32)
Δ = 128
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*1}=\frac{0-8\sqrt{2}}{2} =-\frac{8\sqrt{2}}{2} =-4\sqrt{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*1}=\frac{0+8\sqrt{2}}{2} =\frac{8\sqrt{2}}{2} =4\sqrt{2} $
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