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8k^2-13=0
a = 8; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·8·(-13)
Δ = 416
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{26}}{2*8}=\frac{0-4\sqrt{26}}{16} =-\frac{4\sqrt{26}}{16} =-\frac{\sqrt{26}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{26}}{2*8}=\frac{0+4\sqrt{26}}{16} =\frac{4\sqrt{26}}{16} =\frac{\sqrt{26}}{4} $
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