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8k^2=128
We move all terms to the left:
8k^2-(128)=0
a = 8; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·8·(-128)
Δ = 4096
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4096}=64$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*8}=\frac{-64}{16} =-4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*8}=\frac{64}{16} =4 $
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