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4k^2+16k+4=0
a = 4; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·4·4
Δ = 192
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{3}}{2*4}=\frac{-16-8\sqrt{3}}{8} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{3}}{2*4}=\frac{-16+8\sqrt{3}}{8} $
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