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