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4k^2-13=0
a = 4; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·4·(-13)
Δ = 208
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*4}=\frac{0-4\sqrt{13}}{8} =-\frac{4\sqrt{13}}{8} =-\frac{\sqrt{13}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*4}=\frac{0+4\sqrt{13}}{8} =\frac{4\sqrt{13}}{8} =\frac{\sqrt{13}}{2} $
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