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36k^2-16=0
a = 36; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·36·(-16)
Δ = 2304
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{2304}=48$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*36}=\frac{-48}{72} =-2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*36}=\frac{48}{72} =2/3 $
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