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16k^2-256=0
a = 16; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·16·(-256)
Δ = 16384
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{16384}=128$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128}{2*16}=\frac{-128}{32} =-4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128}{2*16}=\frac{128}{32} =4 $
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