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4k^2+18k+18=0
a = 4; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·4·18
Δ = 36
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*4}=\frac{-24}{8} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*4}=\frac{-12}{8} =-1+1/2 $
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