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(K)^2-48=64
We move all terms to the left:
(K)^2-48-(64)=0
We add all the numbers together, and all the variables
K^2-112=0
a = 1; b = 0; c = -112;
Δ = b2-4ac
Δ = 02-4·1·(-112)
Δ = 448
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}$$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*1}=\frac{0-8\sqrt{7}}{2} =-\frac{8\sqrt{7}}{2} =-4\sqrt{7} $$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*1}=\frac{0+8\sqrt{7}}{2} =\frac{8\sqrt{7}}{2} =4\sqrt{7} $
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