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4k^2-80=0
a = 4; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·4·(-80)
Δ = 1280
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*4}=\frac{0-16\sqrt{5}}{8} =-\frac{16\sqrt{5}}{8} =-2\sqrt{5} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*4}=\frac{0+16\sqrt{5}}{8} =\frac{16\sqrt{5}}{8} =2\sqrt{5} $
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